1.8: Improper Functions (Exercises)

[exer:1] Suppose \(g\) and \(h\) are differentiable on \([a,b]\), with

\[a \le g(y) \le b \text{\quad and\quad} a \le h(y) \le b, \quad c \le y \le d.\]

Let \(f\) and \(f_{y}\) be continuous on \([a,b]\times [c,d]\). Derive Liebniz’s rule:

\[\begin{aligned} \frac{d}{dy}\int_{g(y)}^{h(y)}f(x,y)\,dx &=&f(h(y),y)h'(y)-f(g(y),y)g'(y)\\&&+\int_{g(y)}^{h(y)}f_{y}(x,y)\,dx.\end{aligned}\]

(Hint: Define \(H(y,u,v)=\int_{u}^{v}f(x,y)\,dx\) and use the chain rule.)

[exer:2] Adapt the proof of Theorem [theorem:2] to prove Theorem [theorem:3].

[exer:3] Adapt the proof of Theorem [theorem:4] to prove Theorem [theorem:5].

[exer:4] Show that Definition [definition:3] is independent of \(c\); that is, if \(\int_{a}^{c}f(x,y)\,dx\) and \(\int_{c}^{b}f(x,y)\,dx\) both converge uniformly on \(S\) for some \(c\in (a,b)\), then they both converge uniformly on \(S\) and every \(c\in (a,b)\).

[exer:5]

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Show that if \(f\) is bounded on \([a,b]\times [c,d]\) and \(\int_{a}^{b}f(x,y)\,dx\) exists as a proper integral for each \(y\in [c,d]\), then it converges uniformly on \([c,d]\) according to all of Definition [definition:1]–[definition:3].

Give an example to show that the boundedness of \(f\) is essential in (a).

[exer:6] Working directly from Definition [definition:1], discuss uniform convergence of the following integrals:

(a) \(\displaystyle{\int_{0}^{\infty}\frac{dx}{1+y^{2}x^{2}}\,dx}\)	(b) \(\displaystyle{\int_{0}^{\infty}e^{-xy}x^{2}\,dx}\)
(c) \(\displaystyle{\int_{0}^{\infty}x^{2n}e^{-yx^{2}}\,dx}\)	(d) \(\displaystyle{\int_{0}^{\infty}\sin xy^{2}\,dx}\)
(e) \(\displaystyle{\int_{0}^{\infty}(3y^{2}-2xy)e^{-y^{2}x}\,dx}\)	(f) \(\displaystyle{\int_{0}^{\infty}(2xy-y^{2}x^{2})e^{-xy}\,dx}\)

[exer:7] Adapt the proof of Theorem [theorem:6] to prove Theorem [theorem:7].

[exer:8] Use Weierstrass’s test to show that the integral converges uniformly on \(S:\)

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\(\displaystyle{\int_{0}^{\infty}e^{-xy}\sin x\,dx}\), \(S=[\rho,\infty)\),\(\rho>0\)

\(\displaystyle{\int_{0}^{\infty}\displaystyle{\frac{\sin x}{x^{y}}}\,dx}\), \(S=[c,d]\), \(1<c<d<2\)

\(\displaystyle{\int_{1}^{\infty}e^{-px}\displaystyle{\frac{\sin xy}{x}}\,dx}\), \(p>0\), \(S=(-\infty,\infty)\)

\(\displaystyle{\int_{0}^{1}\frac{e^{xy}}{(1-x)^{y}}}\,dx\), \(S=(-\infty,b)\),\(b<1\)

\(\displaystyle{\int_{-\infty}^{\infty}\frac{\cos xy}{1+x^{2}y^{2}}}\,dx\), \(S=(-\infty,-\rho]\cup[\rho,\infty)\),\(\rho>0\).

\(\displaystyle{\int_{1}^{\infty}e^{-x/y}\,dx}\), \(S=[\rho,\infty)\),\(\rho>0\)

\(\displaystyle{\int_{-\infty}^{\infty}e^{xy}e^{-x^{2}}\,dx}\), \(S=[-\rho,\rho]\),\(\rho>0\)

\(\displaystyle{\int_{0}^{\infty}\frac{\cos xy-\cos ax}{x^{2}}\,dx}\), \(S=(-\infty,\infty)\)

\(\displaystyle{\int_{0}^{\infty}x^{2n}e^{-yx^{2}}\,dx}\), \(S=[\rho,\infty)\),\(\rho>0\), \(n=0\), \(1\), \(2\),…

[exer:9]

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Show that

\[\Gamma(y)=\int_{0}^{\infty} x^{y-1}e^{-x}\,dx\]

converges if \(y>0\), and uniformly on \([c,d]\) if \(0<c<d<\infty\).

Use integration by parts to show that

\[\Gamma(y)=\frac{\Gamma(y+1)}{y},\quad y \ge 0,\]

and then show by induction that

\[\Gamma(y)=\frac{\Gamma(y+n)}{y(y+1)\cdots(y+n-1)}, \quad y>0, \quad n=1,2,3, \dots.\]

How can this be used to define \(\Gamma(y)\) in a natural way for all \(y\ne0\), \(-1\), \(-2\), …? (This function is called the gamma function.)

Show that \(\Gamma(n+1)=n!\) if \(n\) is a positive integer.

Show that

\[\int_{0}^{\infty}e^{-st}t^{\alpha}\,dt =s^{-\alpha-1}\Gamma(\alpha+1), \quad \alpha>-1, \quad s>0.\]

[exer:10] Show that Theorem [theorem:8] remains valid with assumption (c) replaced by the assumption that \(|g_{x}(x,y)|\) is monotonic with respect to \(x\) for all \(y\in S\).

[exer:11] Adapt the proof of Theorem [theorem:8] to prove Theorem [theorem:9].

[exer:12] Use Dirichlet’s test to show that the following integrals converge uniformly on \(S=[\rho,\infty)\) if \(\rho>0\):

(a) \(\displaystyle{\int_{1}^{\infty}\frac{\sin xy}{x^{y}}\,dx}\)	(b) \(\displaystyle{\int_{2}^{\infty}\frac{\sin xy}{\log x}\,dx}\)
(c) \(\displaystyle{\int_{0}^{\infty}\frac{\cos xy}{x+y^{2}}\,dx}\)	(d) \(\displaystyle{\int_{1}^{\infty}\frac{\sin xy}{1+xy}\,dx}\)

[exer:13] Suppose \(g,\) \(g_{x}\) and \(h\) are continuous on \([a,b)\times S,\) and denote \(H(x,y)=\int_{a}^{x}h(u,y)\,du,\) \(a\le x<b.\) Suppose also that

\[\lim_{x\to b-} \left\{\sup_{y\in S}|g(x,y)H(x,y)|\right\}=0 \text{\quad and \quad}\int_{a}^{b}g_{x}(x,y)H(x,y)\,dx\]

converges uniformly on \(S.\) Show that \(\int_{a}^{b}g(x,y)h(x,y)\,dx\) converges uniformly on \(S\).

[exer:14] Prove Theorem [theorem:10] for the case where \(f=f(x,y)\) is continuous on \((a,b]\times [c,d]\).

[exer:15] Prove Theorem [theorem:11] for the case where \(f=f(x,y)\) is continuous on \((a,b]\times [c,d]\).

[exer:16] Show that

\[C(y)=\int_{-\infty}^{\infty}f(x)\cos xy\,dx \text{\quad and\quad} S(y)=\int_{-\infty}^{\infty}f(x)\sin xy\,dx\]

are continuous on \((-\infty,\infty)\) if

\[\int_{-\infty}^{\infty}|f(x)|\,dx<\infty.\]

[exer:17] Suppose \(f\) is continuously differentiable on \([a,\infty)\), \(\lim_{x\to\infty}f(x)=0\), and

\[\int_{a}^{\infty}|f'(x)|\,dx<\infty.\]

Show that the functions

\[C(y)=\int_{a}^{\infty}f(x)\cos xy\,dx \text{\quad and\quad} S(y)=\int_{a}^{\infty}f(x)\sin xy\,dx\]

are continuous for all \(y\ne0\). Give an example showing that they need not be continuous at \(y=0\).

[exer:18] Evaluate \(F(y)\) and use Theorem [theorem:11] to evaluate \(I\):

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\(F(y)=\displaystyle{\int_{0}^{\infty}\frac{dx}{1+y^{2}x^{2}}}\), \(y\ne0\); \(I=\displaystyle{\int_{0}^{\infty}\frac{\tan^{-1}ax-\tan^{-1}bx}{x}\,dx}\), \(a\), \(b>0\)

\(F(y)=\displaystyle{\int_{0}^{\infty}x^{y}\,dx}\), \(y>-1\); \(I=\displaystyle{\int_{0}^{\infty}\frac{x^{a}-x^{b}}{\log x}\,dx}\), \(a\), \(b>-1\)

\(F(y)=\displaystyle{\int_{0}^{\infty}e^{-xy}\cos x\,dx}\), \(y>0\)

\(I=\displaystyle{\int_{0}^{\infty}\frac{e^{-ax}-e^{-bx}}{x}\cos x\,dx}\), \(a\), \(b>0\)

\(F(y)=\displaystyle{\int_{0}^{\infty}e^{-xy}\sin x\,dx}\), \(y>0\)

\(I=\displaystyle{\int_{0}^{\infty}\frac{e^{-ax}-e^{-bx}}{x}\sin x\,dx}\), \(a\), \(b>0\)

\(F(y)=\displaystyle{\int_{0}^{\infty}e^{-x}\sin xy\,dx}\);  \(I=\displaystyle{\int_{0}^{\infty}e^{-x}\displaystyle\frac{1-\cos ax}{x}}\,dx\)

\(F(y)=\displaystyle{\int_{0}^{\infty}e^{-x}\cos xy\,dx}\);  \(I=\displaystyle{\int_{0}^{\infty}e^{-x}\displaystyle\frac{\sin ax}{x}}\,dx\)

[exer:19] Use Theorem [theorem:11] to evaluate:

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\(\displaystyle{\int_{0}^{1}(\log x)^{n}x^{y}\,dx}\), \(y>-1\),\(n=0\), \(1\), \(2\),….

\(\displaystyle{\int_{0}^{\infty}\displaystyle{\frac{dx}{(x^{2}+y)^{n+1}}}\,dx}\), \(y>0\), \(n=0\), \(1\), \(2\), ….

\(\displaystyle{\int_{0}^{\infty}x^{2n+1}e^{-yx^{2}}\,dx}\), \(y>0\), \(n=0\), \(1\), \(2\),….

\(\displaystyle{\int_{0}^{\infty}xy^{x}\,dx}\), \(0<y<1\).

[exer:20]

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Use Theorem [theorem:11] and integration by parts to show that

\[F(y)=\int_{0}^{\infty}e^{-x^{2}}\cos 2xy\,dx\]

satisfies

\[F'+2y F=0.\]

Use part (a) to show that

\[F(y)=\frac{\sqrt{\pi}}{2} e^{-y^{2}}.\]

[exer:21] Show that

\[\int_{0}^{\infty}e^{-x^{2}}\sin 2xy\,dx =e^{-y^{2}}\int_{0}^{y} e^{u^{2}}\,du.\]

(Hint: See Exercise [exer:20].)

[exer:22] State a condition implying that

\[C(y)=\int_{a}^{\infty}f(x)\cos xy\,dx \text{\quad and\quad} S(y)=\int_{a}^{\infty}f(x)\sin xy\,dx\]

are \(n\) times differentiable on for all \(y\ne0\). (Your condition should imply the hypotheses of Exercise [exer:16].)

[exer:23] Suppose \(f\) is continuously differentiable on \([a,\infty)\),

\[\int_{a}^{\infty}|(x^{k}f(x))'|\,dx<\infty,\quad 0\le k\le n,\]

and \(\lim_{x\to\infty}x^{n}f(x)=0\). Show that if

\[C(y)=\int_{a}^{\infty}f(x)\cos xy\,dx \text{\quad and\quad} S(y)=\int_{a}^{\infty}f(x)\sin xy\,dx,\]

then

\[C^{(k)}(y)=\int_{a}^{\infty}x^{k}f(x)\cos xy\,dx \text{\quad and\quad} S^{(k)}(y)=\int_{a}^{\infty}x^{k}f(x)\sin xy\,dx,\]

\(0\le k\le n\).

[exer:24] Differentiating

\[F(y)=\int_{1}^{\infty}\cos\frac{y}{x}\,dx\]

under the integral sign yields

\[-\int_{1}^{\infty}\frac{1}{x}\sin\frac{y}{x}\,dx,\]

which converges uniformly on any finite interval. (Why?) Does this imply that \(F\) is differentiable for all \(y\)?

[exer:25] Show that Theorem [theorem:11] and induction imply Eq. [eq:30].

[exer:26] Prove Theorem [theorem:12].

[exer:27] Show that if \(F(s)=\int_{0}^{\infty}e^{-sx}f(x)\,dx\) converges for \(s=s_{0}\), then it converges uniformly on \([s_{0},\infty)\). (What’s the difference between this and Theorem [theorem:13]?)

[exer:28] Prove: If \(f\) is continuous on \([0,\infty)\) and \(\int_{0}^{\infty}e^{-s_{0}x}f(x)\,dx\) converges, then

\[\lim_{s\to s_{0}+}\int_{0}^{\infty}e^{-sx}f(x)\,dx= \int_{0}^{\infty}e^{-s_{0}x}f(x)\,dx.\]

(Hint: See the proof of Theorem 4.5.12, p. 273.)

[exer:29] Under the assumptions of Exercise [exer:28], show that

\[\lim_{s\to s_{0}+}\int_{r}^{\infty}e^{-sx}f(x)\,dx= \int_{r}^{\infty}e^{-s_{0}x}f(x)\,dx,\quad r>0.\]

[exer:30] Suppose \(f\) is continuous on \([0,\infty)\) and

\[F(s)=\int_{0}^{\infty}e^{-sx}f(x)\,dx\]

converges for \(s = s_{0}\). Show that \(\lim_{s\to\infty}F(s)=0\). (Hint: Integrate by parts.)

[exer:31]

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Starting from the result of Exercise [exer:18](d), let \(b\to\infty\) and invoke Exercise [exer:30] to evaluate

\[\int_{0}^{\infty}e^{-ax} \frac{\sin x}{x}\,dx, \quad a>0.\]

Use (a) and Exercise [exer:28] to show that

\[\int_{0}^{\infty} \frac{\sin x}{x}\,dx =\frac{\pi}{2}.\]

[exer:32]

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Suppose \(f\) is continuously differentiable on \([0,\infty)\) and

\[|f(x)| \le Me^{s_{0}x}, \quad 0\le x\le \infty.\]

Show that

\[G(s)=\int_{0}^{\infty} e^{-sx}f'(x)\,dx\]

converges uniformly on \([s_{1},\infty)\) if \(s_{1}>s_{0}\). (Hint: Integrate by parts.)

Show from part (a) that

\[G(s)=\int_{0}^{\infty} e^{-sx}xe^{x^{2}}\sin e^{x^{2}}\,dx\]

converges uniformly on \([\rho,\infty)\) if \(\rho>0\). (Notice that this does not follow from Theorem [theorem:6] or [theorem:8].)

[exer:33] Suppose \(f\) is continuous on \([0,\infty)\),

\[\lim_{x\to0+}\frac{f(x)}{x}\]

exists, and

\[F(s)=\int_{0}^{\infty}e^{-sx}f(x)\,dx\]

converges for \(s=s_{0}\). Show that

\[\int_{s_{0}}^{\infty}F(u)\,du=\int_{0}^{\infty}e^{-s_{0}x}\frac{f(x)}{x}\,dx.\]

Answers to selected exercises

If \(f(x,y)=1/y\) for \(y\ne0\) and \(f(x,0)=1\), then \(\int_{a}^{b}f(x,y)\,dx\) does not converge uniformly on \([0,d]\) for any \(d>0\).

, (d), and (e) converge uniformly on \((-\infty,\rho]\cup[\rho,\infty)\) if \(\rho>0\);  (b), (c), and (f) converge uniformly on \([\rho,\infty)\) if \(\rho>0\).

Let \(C(y)=\displaystyle{\int_{1}^{\infty}\frac{\cos xy}{x}\,dx}\) and \(S(y)=\displaystyle{\int_{1}^{\infty}\frac{\sin xy}{x}\,dx}\). Then \(C(0)=\infty\) and \(S(0)=0\), while \(S(y)=\pi/2\) if \(y\ne0\).

\(F(y)=\displaystyle{\frac{\pi}{2|y|}}\);\(I=\displaystyle{\frac{\pi}{2}\log\frac{a}{b}}\) \(F(y)=\displaystyle{\frac{1}{y+1}}\);\(I=\displaystyle{\log\frac{a+1}{b+1}}\)

\(F(y)=\displaystyle{\frac{y}{y^{2}+1}}\); \(I=\displaystyle{\frac{1}{2}\,\frac{b^{2}+1}{a^{2}+1}}\)

(d) \(F(y)=\displaystyle{\frac{1}{y^{2}+1}}\);\(I=\tan^{-1}b-\tan^{-1}a\)

(e) \(F(y)=\displaystyle{\frac{y}{y^{2}+1}}\);\(I=\displaystyle{\frac{1}{2}}\log(1+a^{2})\)

(f) \(F(y)=\displaystyle{\frac{1}{y^{2}+1}}\);\(I=\tan^{-1}a\)

\((-1)^{n}n!(y+1)^{-n-1}\) \(\pi2^{-2n-1}\displaystyle{\binom{2n}{n}}y^{-n-1/2}\)

(c) \(\displaystyle{\frac{n!}{2y^{n+1}}}\) \((\log y)^{-2}\) (d) \(\displaystyle{\frac{1}{(\log x)^{2}}}\)

\(\displaystyle{\int_{-\infty}^{\infty}|x^{n}f(x)|\,dx<\infty}\)

No; the integral defining \(F\) diverges for all \(y\).

  \(\displaystyle{\frac{\pi}{2}}-\tan^{-1}a\)

Beginning of manual

1. If \(H(y,u,v)=\displaystyle{\int_{u}^{v}f(x,y)\,dx}\) then

\[H_{u}(y,u,v)=-f(u,y), \quad H_{v}(y,u,v)=f(v,y),\]

and, by Theorem 1, \(H_{y}(u,v,y) =\displaystyle{\int_{u}^{v}f_{y}(x,y)\,dx}\). If

\[F(y)=H(y,g(y),h(y))=\int_{g(y)}^{h(y)}f(x,y)\,dx,\]

then

\[\begin{aligned} F'(y)&=&H_{v}(y, g(y),h(y))h'(y)+H_{u}(y,g(y),h(y))g'(y)+ H_{y}(y,g(y),h(y))\\ &=& f(h(y),y)h'(y)-f(g(y),y)g'(y) +\int_{g(y)}^{h(y)} f_{y}(x,y)\,dx.\end{aligned}\]

2. Theorem 3 (Cauchy Criterion for Convergence of an Improper Integral II) Suppose \(g\) is integrable on every finite closed subinterval of \((a,b]\) and denote

\[G(r)=\int_{r}^{b}g(x)\,dx,\quad a< r\le b.\]

Then the improper integral \(\int_{a}^{b}g(x)\,dx\) converges if and only if\(,\) for each \(\epsilon >0,\) there is an \(r_{0}\in(a,b]\) such that

\[\tag{A} |G(r)-G(r_{1})|\le\epsilon,\quad a<r,r_{1}\le r_{0}.\]

For necessity, suppose \(\int_{a}^{b}g(x)\,dx=L\). By definition, this means that for each \(\epsilon>0\) there is an \(r_{0}\in (a,b]\) such that

\[|G(r)-L|<\frac{\epsilon}{2} \text{\quad and\quad} |G(r_{1})-L|<\frac{\epsilon}{2}, \quad a< r,r_{1}\le r_{0}.\]

Therefore,

\[\begin{aligned} |G(r)-G(r_{1})|&=&|(G(r)-L)-(G(r_{1})-L)|\\ &\le& |G(r)-L|+|G(r_{1})-L|\le \epsilon,\quad a< r,r_{1}\le r_{0}.\end{aligned}\]

For sufficiency, (A) implies that

\[|G(r)|= |G(r_{1})+(G(r)-G(r_{1}))|\le |G(r_{1})|+|G(r)-G(r_{1})|\le |G(r_{1})|+\epsilon,\]

\(a< r_{1}\le r_{0}\). Since \(G\) is also bounded on the compact set \([r_{0},b]\) (Theorem 5.2.11, p. 313), \(G\) is bounded on \((a,b]\). Therefore the monotonic functions

\[\overline{G}(r)=\sup\left\{G(r_{1})\, \big|\, a<r_{1}\le r\right\} \text{\quad and\quad} \underline{G}(r)=\inf\left\{G(r_{1})\, \big|\, a<r_{1}\le r\right\}\]

are well defined on \((a,b]\), and

\[\lim_{r\to a+}\overline{G}(r)=\overline{L} \text{\quad and\quad} \lim_{r\to a+}\underline{G}(r)=\underline{L}\]

both exist and are finite (Theorem 2.1.11, p. 47). From (A),

\[\begin{aligned} |G(r)-G(r_{1})|&=&|(G(r)-G(r_{0}))-(G(r_{1})-G(r_{0}))|\\ &\le &|G(r)-G(r_{0})|+|G(r_{1})-G(r_{0})|\le 2\epsilon,\end{aligned}\]

so \(\overline{G}(r)-\underline{G}(r)\le 2\epsilon\). Since \(\epsilon\) is an arbitrary positive number, this implies that

\[\lim_{r\to a+}(\overline{G}(r)-\underline{G}(r))=0,\]

so \(\overline{L}=\underline{L}\). Let \(L=\overline{L}=\underline{L}\). Since

\[\underline{G}(r)\le G(r)\le \overline{G}(r),\]

it follows that \(\lim_{r\to a+} G(r)=L\).

3. Theorem 5 (Cauchy Criterion for Uniform Convergence II) The improper integral

\[\int_{a}^{b}f(x,y)\,dx =\lim_{r\to a+}\int_{r}^{b}f(x,y)\,dx\]

converges uniformly on \(S\) if and only if\(,\) for each \(\epsilon>0,\) there is an \(r_{0}\in (a,b]\) such that

\[\tag{A} \left|\int_{r_{1}}^{r}f(x,y)\,dx\right|< \epsilon, \quad y\in S, \quad a <r,r_{1}\le r_{0}.\]

Suppose \(\int_{a}^{b} f(x,y)\,dx\) converges uniformly on \(S\) and \(\epsilon>0\). From Definition 2, there is an \(r_{0}\in (a,b]\) such that

\[\tag{B} \left|\int_{a}^{r}f(x,y)\,dx\right| <\frac{\epsilon}{2} \text{\quad and\quad} \left|\int_{a}^{r_{1}}f(x,y)\,dx\right| <\frac{\epsilon}{2},\, y\in S, \, a< r,r_{1}\le r_{0}.\]

Since

\[\int_{r_{1}}^{r}f(x,y)\,dx= \int_{r_{1}}^{b}f(x,y)\,dx- \int_{r}^{b}f(x,y)\,dx\]

(B) and the triangle inequality imply (A).

For the converse, denote

\[F(y)=\int_{r}^{b}f(x,y)\,dx.\]

Since (A) implies that

\[|F(r,y)-F(r_{1},y)|\le \epsilon, \quad y\in S, \quad a< r, r_{1}\le r_{0},\]

Theorem 2 with \(G(r)=F(r,y)\) (\(y\) fixed but arbitrary in \(S\)) implies that \(\int_{a}^{b} f(x,y)\,dx\) converges pointwise for \(y\in S\). Therefore, if \(\epsilon>0\) then, for each \(y\in S\), there is an \(r_{0}(y) \in (a,b]\) such that

\[\tag{C} \left|\int_{a}^{r}f(x,y)\,dx\right|\le \epsilon, \quad y\in S,\quad a<r\le r_{0}(y).\]

For each \(y\in S\), choose \(r_{1}(y)\le \min[{r_{0}(y),r_{0}}]\). Then

\[\int_{a}^{r}f(x,y)\,dx = \int_{a}^{r_{1}(y)}f(x,y)\,dx+ \int_{r_{1}(y)}^{r}f(x,y)\,dx, \quad\]

so (A), (C), and the triangle inequality imply that

\[\left|\int_{a}^{r} f(x,y)\,dx\right|\le 2\epsilon,\quad y\in S,\quad a<r\le r_{0}\]

4. From Definition 3, \(\int_{a}^{b}f(x,y)\,dx\) converges uniformly on \(S\) if and only if \(\int_{a}^{c}f(x,y)\,dx\) and \(\int_{c}^{b}f(x,y)\,dx\) both converge uniformly on \(S\), where \(c\in(a,b)\). From Theorems 4 and Theorem 5, this is true if and only if, for any \(\epsilon>0\) there are points \(r_{0}\) and \(s_{0}\) in \((a,b)\) such that

\[\left|\int_{r}^{r_{1}}f(x,y)\,dx\right|\le \epsilon,\quad y\in S,\quad r_{0}\le r,r_{1}<b\]

and

\[\left|\int_{s_{1}}^{s}f(x,y)\,dx\right|\le \epsilon,\quad y\in S,\quad a< s,s_{1}<s_{0}.\]

These conditions are independent of \(c\).

5. (a) If \(|f(x,y)|\le M\) on \([a,b]\times [c,d]\) then

\[\left|\int_{r_{1}}^{r_{2}}f(x,y)\,dx\right|\le M|r_{2}-r_{1}|\]

so the Cauchy convergence theorems imply the conclusion.

Define \(f=f(x,y)\) on \([0,1]\times [0,1]\) by

\[f(x,y)= \begin{cases}\displaystyle\frac{1}{y} &\text{if\quad} 0<y\le 1,\\ 1&\text{if\quad} y=0. \end{cases}\]

Then

\[\int_{r_{1}}^{r_{2}}f(x,y)\,dx= \begin{cases}\displaystyle\frac{r_{2}-r_{1}}{y} &\text{if\quad} 0<y\le 1,\\ r_{2}-r_{1}&\text{if\quad} y=0. \end{cases}\]

Therefore \(f\) does not satisfy the requirements of Cauchy’s convergence theorems.

6. In all parts \(I(y)\) denotes the given integral.

\(I(0)=\infty\). If \(y\ne0\) let \(u=xy\); then \(I(y)=\displaystyle{\frac{1}{y}\int_{0}^{\infty}\frac{du}{1+u^{2}}}\). If \(\rho>0\) and \(\epsilon >0\), choose \(r\) so that \(\displaystyle{\int_{r}^{\infty}\frac{du}{1+u^{2}}}< \rho\epsilon\). Then \(\displaystyle{\frac{1}{|y|}\int_{r}^{\infty}\frac{du}{1+u^{2}}}<\epsilon\) if \(|y|\ge \rho\), so \(I(y)\) converges uniformly on \((-\infty, \rho]\bigcup [\rho,\infty)\) if \(\rho>0\).

\(I(y)=\infty\) if \(y\le0\). If \(y>0\) let \(u=xy\); then \(I(y)=\displaystyle{\frac{1}{y^{3}}\int_{0}^{\infty}e^{-u}u^{2}\,du}\). If \(\rho>0\) and \(\epsilon >0\), choose \(r\) so that \(\displaystyle{\int_{r}^{\infty}e^{-u}u^{2}\,du}<\rho^{3}\epsilon\). Then \(\displaystyle{\frac{1}{y^{3}}\int_{r}^{\infty}e^{-u}u^{2}\,du}<\epsilon\) if \(y\ge \rho\), so \(I(y)\) converges uniformly on \([\rho,\infty)\) if \(\rho>0\).

\(I(y)=\infty\) if \(y\le0\). If \(y>0\) let \(u=xy^{1/2}\); then \(I(y)=\displaystyle{y^{-n-1/2}\int_{0}^{\infty}u^{2n}e^{-u}\,du}\). If \(\rho>0\) and \(\epsilon >0\), we can choose \(r\) so that \(\displaystyle{\int_{r}^{\infty}u^{2n}e^{-u}\,du}<\epsilon \rho^{n+1/2}\). Then \(y^{-n-1/2}\displaystyle{\int_{r}^{\infty}u^{2n}e^{-u}\,du}<\epsilon\) if \(y\ge \rho\), so \(I(y)\) converges uniformly on \(S=[\rho,\infty)\) if \(\rho>0\).

Since \(I(-y)=-I(y)\), it suffices to assume that \(y>0\). If \(u=yx^{2}\) then \(I(y)=\displaystyle{\frac{1}{2\sqrt{y}}\int_{0}^{\infty}\frac{\sin u\,du}{\sqrt{u}}}\). From Example 3.4.14 (p. 162), this integral converges conditionally. If \(\rho>0\) and \(\epsilon >0\), we can choose \(r\) so that \(\displaystyle{\left|\int_{r}^{\infty}\frac{\sin u\,du}{\sqrt{u}}\right|}<2\epsilon\sqrt{\rho}\), so \(I(y)\) converges uniformly on \((-\infty,-\rho]\bigcup[\rho,\infty)\) if \(\rho>0.\)

If \(u=y^{2}x\) then \(\displaystyle{I(y)=3\int_{0}^{\infty}e^{-u}\,du -\frac{2}{y^{3}}\int_{0}^{\infty} ue^{-u}\,du}\). If \(\rho>0\), we can choose \(r\) so that \(\displaystyle{3\int_{r}^{\infty}e^{-u}\,du<\frac{\epsilon}{2}}\) and \(\displaystyle{\int_{r}^{\rho}ue^{-u}\,du<\frac{\rho^{3}\epsilon}{2}}\). Then

\[\left|3\int_{r}^{\infty}e^{-u}\,du -\frac{3}{y^{3}}\int_{r}^{\infty} ue^{-u}\,du\right|<\epsilon \text{\quad if\quad} |y|\ge \rho,\]

so \(I(y)\) converges uniformly on \((-\infty, \rho]\bigcup [\rho,\infty)\) if \(\rho>0\).

\(I(y)=-\infty\) if \(y\le0\). If \(y>0\), let \(u=xy\); then \(I(y)=\displaystyle{\frac{1}{y}\int_{0}^{\infty}(2u-u^{2})e^{-u}\,du}\). If \(\rho>0\) and \(\epsilon >0\), we can choose \(r\) so that \(\displaystyle{\int_{r}^{\infty}|2u-u^{2}|e^{-u}\,du}<\epsilon \rho\), so \(I(y)\) converges uniformly on \(S=[\rho,\infty)\) if \(\rho>0\).

7. Theorem 7 (Weierstrass’s Test for Absolute Uniform Convergence II) Suppose \(f=f(x,y)\) is locally integrable \((a,b]\) and, for some \(b_{0}\in (a,b],\)

\[\tag{A} |f(x,y)| \le M(x), \: y\in S, \: x\in (a,b_{0}],\]

where

\[\int_{a}^{b_{0}}M(x)\,dx<\infty.\]

Then \(\int_{a}^{b}f(x,y)\,dx\) converges absolutely uniformly on \(S.\)

Denote \(\int_{a}^{b_{0}}M(x)\,dx=L<\infty\). By definition, for each \(\epsilon>0\) there is an \(r_{0}\in (a,b_{0}]\) such that

\[L-\epsilon \le \int_{r}^{b_{0}}M(x)\,dx \le L,\quad a<r\le r_{0}.\]

Therefore, if \(a<r_{1}< r\le r_{0}\), then

\[0\le \int_{r_{1}}^{r}M(x)\,dx=\left(\int_{r_{1}}^{b_{0}}M(x)\,dx -L\right)- \left(\int_{r}^{b_{0}}M(x)\,dx -L\right)<\epsilon.\]

This and (A) imply that

\[\int_{r_{1}}^{r}|f(x,y)|\,dx\le \int_{r_{1}}^{r} M(x)\,dx <\epsilon, \: y\in S,\: a<r_{1}\le r_{0}\le b.\]

Now Theorem 5 implies the stated conclusion.

\(|e^{-xy}\sin x|\le e^{-\rho x}\) if \(y\ge\rho\) and \(\int_{\rho}^{\infty}e^{-\rho x}\,dx<\infty\).

\(\displaystyle{\int_{0}^{\infty}\frac{\sin x\,dx}{x^{y}}}=I_{1}(y)+I_{2}(y)\), where

\[I_{1}(y)=\int_{0}^{1}\frac{\sin x\,dx}{x^{y}} \text{\quad and\quad} I_{2}(y)=\int_{1}^{\infty}\frac{\sin x\,dx}{x^{y}}\]

are both improper integrals. Since

\[\sin x= x-\left(\frac{x^{3}}{3!}-\frac{x^{5}}{5!}\right) -\left(\frac{x^{7}}{7!}-\frac{x^{9}}{9!}\right)+ \cdots <x,\quad 0\le x \le 1.\]

If \(0\le 1\) and \(y\le d\le 2\), then

\[\left|\frac{\sin x}{x^{y}}\right|\le x^{1-y}\le x^{1-d} \text{\:so\:} \int_{0}^{1}x^{-1+y}\,dx<\int_{0}^{1}x^{-1+d}\,dx=\frac{1}{2-d}\]

so \(I_{1}(y)\) converges absolutely uniformly on \(S\). Since \(c>1\),

\[\frac{|\sin x|}{x^{y}}\le x^{-c} \text{\quad and\quad} \int_{1}^{\infty}x^{-c} \,dx=\frac{1}{c-1}\text{\quad if \quad}\]

\(I_{2}(y)\) converges absolutely uniformly on \(S\).

If \(x\ge 1\) then \(\displaystyle{e^{-px}\left|\frac{\sin xy}{x}\right|\le e^{-px}}\) for all \(y\) and \(\displaystyle{\int_{1}^{\infty}e^{-px}\,dx<\infty}\), since \(p>0\).

\(\displaystyle{\frac{e^{xy}}{(1-x)^{y}}}\le \displaystyle{\frac{e^{b}}{(1-x)^{b}}}\), if \(0\le x<1\) and \(y\le b\), and \(\displaystyle{\int_{0}^{1}(1-x)^{-b}\,dx}<\infty\) if \(b<1\).

If \(|y|\ge \rho>0\) then \(\displaystyle{\left|\frac{\cos xy}{1+x^{2}y^{2}}\right|\le \frac{1}{1+\rho^{2}x^{2}}}\) for all \(x\), and \(\displaystyle{\int_{0}^{\infty}\frac{dx}{1+\rho^{2}x^{2}}}<\infty\).

If \(y\ge \rho>0\) then \(e^{-x/y}\le e^{-x/\rho}\) and \(\displaystyle{\int_{0}^{\infty}e^{-x/\rho}\,dx<\infty}\).

If \(|y|\le \rho\) then \(e^{xy}e^{-x^{2}}\le e^{x\rho}e^{-x^{2}}\) and \(\displaystyle{\int_{-\infty}^{\infty}e^{x\rho}e^{-x^{2}}\,dx}<\infty\).

If \(|x|\ge 1\) then \(|\cos xy-\cos ax|\le 2\) and \(\displaystyle{\int_{1}^{\infty}\frac{2\,dx}{x^{2}}}<\infty\).

If \(y\ge \rho>0\) then \(e^{-yx^{2}}\le e^{-\rho x^{2}}\) and \(\displaystyle{\int_{0}^{\infty} x^{2n}e^{-\rho x^{2}}\,dx<\infty}\).

9. (a) If \(0<x<1\) then \(|x^{y-1}e^{-x}|<x^{c-1}\). Therefore, since \(\displaystyle{\int_{0}^{1}x^{c-1}\,dx}<\infty\) if \(c>0\), \(\int_{0}^{1}x^{y-1}e^{-x}\,dx\) converges uniformly on \([c,\infty)\) if \(c>0\), by Theorem 7. If \(x>1\) then \(|x^{y-1}e^{-x}|\le x^{d-1}e^{-x}\) if \(y\le d\). Therefore, since \(\displaystyle{\int_{1}^{\infty}x^{d-1}e^{-x}}\,dx<\infty\), \(\displaystyle{\int_{1}^{\infty}x^{y-1}e^{-x}}\,dx\) converges uniformly on \((-\infty,d]\) for every \(d\), by Theorem 6. Hence, \(\displaystyle{\int_{0}^{\infty}x^{y-1}e^{-x}\,dx}\) converges uniformly on \([c,d]\) if \(c>0\).

If \(y>0\) then

\[\Gamma(y)=\int_{0}^{\infty}x^{y-1}e^{-x}\,dx =\frac{x^{y}e^{-x}}{y}\biggr|_{0}^{\infty}+ \frac{1}{y}\int_{0}^{\infty}x^{y}e^{-x}\,dx =\frac{\Gamma(y+1)}{y}. \tag{A}\]

Therefore

\[\Gamma(y)=\frac{\Gamma(y+n)}{y(y+1)\cdots (y+n-1)} \tag{B}\]

is true when \(n=1\). Now suppose it is true for given positive integer \(n\), and replace \(y\) by \(y+n\) in (A):

\[\Gamma(y+n)=\frac{\Gamma(y+n+1)}{y+n}.\]

Substituting this into (B) yields

\[\Gamma(y)=\frac{\Gamma(y+n+1)}{y(y+1)\cdots (y+n)},\]

which completes the induction.

If \(-n <y<-n+1\) with \(n\le 1\) then \(0<y+n<1\) and we can compute \(\Gamma(y+n)\) from the definition in part (a):

\[\Gamma(y+n)=\int_{0}^{\infty}x^{y+n-1}e^{-x}\,dx.\]

Then we can define \(\Gamma(y)\) by (B).

The assertion is true if \(n=1\), since

\[\Gamma(2)=\int_{0}^{\infty}xe^{-x}\,dx = -xe^{-x}\biggr|_{0}^{\infty}+\int_{0}^{\infty} e^{-x}\,dx = 1.\]

If \(\Gamma(n+1)=n!\) for some \(n\ge 1\), then

\[\begin{aligned} \Gamma(n+2)&=&\int_{0}^{\infty}x^{n+1}e^{-x}\,dx= -e^{-x} x^{n+1}\biggr|_{0}^{\infty}+(n+1) \int_{0}^{\infty}x^{n+1}e^{-x}\,dx\\ &=&(n+1)\Gamma(n+1)=(n+1)n!=(n+1)!,\end{aligned}\]

which completes the induction proof.

The change of variable \(x=st\) yields

\[\int_{0}^{\infty}e^{-st}t^{\alpha}\,dt=\frac{1}{s^{\alpha+1}} \int_{0}^{\infty}x^{\alpha}e^{-x}\,dx=\frac{1}{s^{\alpha+1}} \Gamma(\alpha+1),\]

from the definition of the Gamma function.

Since \(|g_{x}(x,y)|\) is monotonic with respect to \(x\),

\[\tag{A} \int_{r}^{r_{1}}|g_{x}(x,y)|\,dx=|g(r_{1},y)-g(r,y)|,\quad a\le r<r_{1}<b.\]

From Assumption (a) of Theorem 8, if \(\epsilon>0\) there is an \(r_{0}\in [a,b)\) such that

\[|g(s,y)|\le \epsilon,\quad y\in S,\quad r_{0}\le s<b.\]

Therefore, (A) implies that

\[\int_{r}^{r_{1}}|g_{x}(x,y)\,dx\le 2\epsilon,\quad y\in S,\quad r_{0}\le r\le r_{1}<b.\]

Now Theorem 4 implies that \(\int_{a}^{b}|g(x,y)|\,dx\) converges uniformly on \(S\), which is assumption (c) of Theorem 8.

) If \(g,\) \(g_{x},\) and \(h\) are continuous on \((a,b]\times S\) then

\[\int_{a}^{b}g(x,y)h(x,y)\,dx\]

converges uniformly on \(S\) if the following conditions are satisfied:

()0pt0pt14pt 8pt22pt0pt

\(\displaystyle{\lim_{x\to a+}\left\{\sup_{y\in S}|g(x,y)|\right\}=0};\)

There is a constant \(M\) such that

\[\sup_{y\in S}\left|\int_{x}^{b}h(u,y)\,du\right| \le M, \quad a< x\le b;\]

\(\int_{a}^{b}|g_{x}(x,y)|\,dx\) converges uniformly on \(S\).

If

\[\tag{A} H(x,y)=\int_{x}^{b}h(u,y)\,du\]

then integration by parts yields

\[\begin{aligned} \int_{r_{1}}^{r}g(x,y)h(x,y)\,dx&=&-\int_{r_{1}}^{r}g(x,y)H_{x}(x,y)\,dx \\ &=&-g(r,y)H(r,y)+g(r_{1},y)H(r_{1},y)\\ &&+\int_{r_{1}}^{r}g_{x}(x,y)H(x,y)\,dx.\end{aligned}\]

Therefore, since assumption (b) and (A) imply that \(|H(x,y)|\le M\), \((x,y)\in (a,b]\times S\),

\[\tag{B} \left|\int_{r_{1}}^{r}g(x,y)h(x,y)\,dx\right|\le M\left(2\sup_{a<x\le r} |g(x,y)|+\int_{r_{1}}^{r}|g_{x}(x,y)|\,dx\right)\]

on \([r_{1},r]\times S\). Now suppose \(\epsilon>0\). From assumption (a), there is an \(r_{0} \in [a,b)\) such that \(|g(x,y)|<\epsilon\) on \(S\) if \(a< x \le r_{0} \le b\). From assumption (c) and Theorem 5, there is an \(s_{0}\in (a,b]\) such that

\[\int_{r_{1}}^{r}|g_{x}(x,y)|\,dx<\epsilon,\quad y\in S,\quad a<r_{1}<r\le s_{0}.\]

Therefore (B) implies that

\[\left|\int_{r_{1}}^{r}g(x,y)h(x,y)\right| < 3M\epsilon,\quad y\in S,\quad a<r_{1}<r\min(r_{0},s_{0})\]

Now Theorem 5 implies the stated conclusion.

Denote \(F(y)=\displaystyle{\int_{1}^{\infty}\frac{\sin xy}{x^{y}}}\) and, with \(1\le r< r_{1}\),

\[\begin{aligned} F(r,r_{1},y)=\int_{r}^{r_{1}}\frac{\sin xy}{x^{y}}\,dx &=& -\frac{\cos xy}{yx^{y}}\biggr|_{r}^{r_{1}}- \int_{r}^{r_{1}}\frac{\cos xy}{x^{y+1}}\,dx\\ &=& \frac{\cos r y}{yr^{y}}-\frac{\cos r_{1}y}{yr_{1}^{y}}- \int_{r}^{r_{1}}\frac{\cos xy}{x^{y+1}}\,dx.\end{aligned}\]

Therefore

\[|F(r,r_{1},y)|\le \frac{2}{yr^{y}}+\int_{r}^{r_{1}}x^{-y-1}\,dx<\frac{3}{yr^{y}}, \quad r,y>0.\]

Now Theorem 4 implies that \(F(y)\) converges uniformly on \([\rho,\infty)\) if \(\rho>0\).

(b) Denote \(F(y)=\displaystyle{\int_{2}^{\infty}\frac{\sin xy}{\log x}\,dx}\) and, with \(2\le r< r_{1}\),

\[F(r,r_{1},y)=\int_{r}^{r_{1}}\frac{\sin xy}{\log x}\,dx =-\frac{\cos xy}{y\log x}\biggr|_{r}^{r_{1}}- \frac{1}{y}\int_{r}^{r_{1}}\frac{\cos xy}{x(\log x)^{2}}\,dx.\]

Therefore

\[|F(r,r_{1},y)|\le \frac{1}{y}\left|\frac{2}{\log r}+\int_{r}^{r_{1}} \frac{dx}{x(\log x)^{2}}\right|\le \frac{3}{y\log r}.\]

Now Theorem 4 implies that \(F(y)\) converges uniformly on \([\rho,\infty)\) if \(\rho>0\).

(c) Denote \(F(y)=\displaystyle{\int_{0}^{\infty}\frac{\cos xy}{x+y^{2}}}\), and, with \(0<r<r_{1}\),

\[F(r,r_{1},y)=\int_{r}^{r_{1}}\frac{\cos xy}{x+y^{2}}\,dx= \frac{1}{y}\left(\frac{\sin xy}{x+y^{2}}\biggr|_{r}^{r_{1}} +\int_{r}^{r_{1}}\frac{\sin xy}{(x+y^{2})^{2}}\,dx\right),\]

so

\[|F(r,r_{1},y)|\le \frac{3}{y(r+y^{2})}.\]

Now Theorem 4 implies that \(F(y)\) converges uniformly on \([\rho,\infty)\) if \(\rho>0\).

(d) Denote \(F(y)=\displaystyle{\int_{0}^{\infty}\frac{\sin xy}{1+xy}}\), and, with \(0<r<r_{1}\),

\[F(r,r_{1},y)=\int_{r}^{r_{1}}\frac{\sin xy}{1+xy}\,dx= -\frac{\cos xy}{y(1+xy)}\biggr|_{r}^{r_{1}}-\int_{r}^{r_{1}} \frac{\cos xy}{y^{2}(1+xy)^{2}}\,dx,\]

so

\[|F(r,r_{1},y)|\le \frac{3}{y(1+ry)}.\]

Now Theorem 4 implies that \(F(y)\) converges uniformly on \([\rho,\infty)\) if \(\rho>0\).

13. Integration by parts yields

\[\begin{aligned} \int_{r}^{r_{1}}g(x,y)h(x,y)\,dx&=&\int_{r}^{r_{1}}g(x,y)H_{x}(x,y)\,dx\\ &=&g(r_{1},y)H(r_{1},y)-g(r,y)H(r,y)\\ &&-\int_{r}^{r_{1}}g_{x}(x,y)H(x,y)\,dx,\end{aligned}\]

so

\[\left|\int_{r}^{r_{1}}g(x,y)h(x,y)\,dx\right|\le 2\sup_{x\ge r}\left\{\left\{\sup_{y\in S}|g(x,y)H(x,y)|\right\}\right\}+ \left|\int_{r}^{r_{1}}g_{x}(x,y)H(x,y)\,dx\right|.\]

Now suppose \(\epsilon\ge 0\). From our first assmption, there is an \(s_{0}\in [a,b)\) such that

\[\sup_{x\ge r}\left\{\left\{\sup_{y\in S}|g(x,y)H(x,y)|\right\}\right\}<\epsilon, \quad s_{0}\le r<b.\]

Since \(\int_{a}^{b}g_{x}(x,y)H(x,y)\,dx\) converges uniformly on \(S\), Theorem 4 implies that there is an \(r_{0}\in [a,b)\) such that

\[\left|\int_{r}^{r_{1}}g_{x}(x,y)H(x,y)\,dx\right|\le \epsilon, \quad y\in S, \quad r_{0}\le r<r_{1}<b.\]

Therefore,

\[\left|\int_{r}^{r_{1}}g(x,y)h(x,y)\,dx\right|\le 2\epsilon, \quad y\in S, \quad \max(r_{0},s_{0})\le r<r_{1}<b.\]

Now Theorem 4 implies that \(\int_{a}^{b}g_{x}(x,y)h(x,y)\,dx\) converges uniformly on \(S\).

14. Theorem 10 If \(f=f(x,y)\) is continuous on \((a,b]\times [c,d]\) and

\[\tag{A} F(y)=\int_{a}^{b}f(x,y)\,dx\]

converges uniformly on \([c,d],\) then \(F\) is continuous on \([c,d].\) Moreover\(,\)

\[\tag{B} \int_{c}^{d}\left(\int_{a}^{b}f(x,y)\,dx\right)\,dy =\int_{a}^{b}\left(\int_{c}^{d}f(x,y)\,dy\right)\,dx.\]

We will first show that \(F\) in (A) is continuous on \([c,d]\). Since \(F\) converges uniformly on \([c,d]\), Definition 1 implies that if \(\epsilon>0\), there is an \(r \in (a,b]\) such that

\[\left|\int_{a}^{r}f(x,y)\,dx\right|\le \epsilon, \quad c \le y \le d.\]

Therefore, if \(y\) and \(y_{0}\) are in \([c,d]\), then

\[\begin{aligned} |F(y)-F(y_{0})|&=& \left|\int_{a}^{b}f(x,y)\,dx-\int_{a}^{b}f(x,y_{0})\,dx\right|\\ &\le&\left|\int_{r}^{b}[f(x,y)-f(x,y_{0})]\,dx\right|+ \left|\int_{a}^{r}f(x,y)\,dx\right|\\ &&+\left|\int_{a}^{r}f(x,y_{0})\,dx\right|\\\end{aligned}\]

so

\[\tag{C} |F(y)-F(y_{0})| \le \int_{r}^{b}|f(x,y)-f(x,y_{0})|\,dx +2\epsilon.\]

Since \(f\) is uniformly continuous on the compact set \([r,b]\times [c,d]\) (Corollary 5.2.14, p. 314), there is a \(\delta>0\) such that

\[|f(x,y)-f(x,y_{0})|<\epsilon\]

if \((x,y)\) and \((x,y_{0})\) are in \([r,b]\times [c,d]\) and \(|y-y_{0}|<\delta\). This and (C) imply that

\[|F(y)-F(y_{0})|<(r-a)\epsilon +2\epsilon<(b-a+2)\epsilon\]

if \(y\) and \(y_{0}\) are in \([c,d]\) and \(|y-y_{0}|<\delta\). Therefore \(F\) is continuous on \([c,d]\), so the integral on left side of (B) exists. Denote

\[\tag{D} I= \int_{c}^{d}\left(\int_{a}^{b}f(x,y)\,dx\right)\,dy.\]

We will show that the improper integral on the right side of (B) converges to \(I\). To this end, denote

\[I(r)= \int_{r}^{b}\left(\int_{c}^{d}f(x,y)\,dy\right)\,dx.\]

Since we can reverse the order of integration of the continuous function \(f\) over the rectangle \([r,b]\times [c,d]\) (Corollary 7.2.2, p. 466),

\[I(r)=\int_{c}^{d}\left(\int_{r}^{b}f(x,y)\,dx\right)\,dy.\]

From this and (D),

\[I-I(r)=\int_{c}^{d}\left(\int_{a}^{r}f(x,y)\,dx\right)\,dy.\]

Now suppose \(\epsilon>0\). Since \(\int_{a}^{b}f(x,y)\,dx\) converges uniformly on \([c,d]\), there is an \(r_{0}\in (a,b]\) such that

\[\left|\int_{a}^{r}f(x,y)\,dx\right|<\epsilon, \quad a<r<r_{0}, %###\]

so \(|I-I(r)|<(d-c)\epsilon\), \(a<r<r_{0}\). Hence,

\[\lim_{r\to a+}\int_{r}^{b}\left(\int_{c}^{d}f(x,y)\,dy\right)\,dx= \int_{c}^{d}\left(\int_{a}^{b}f(x,y)\,dx\right)\,dy,\]

which completes the proof of (B).

15. Theorem 11 Let \(f\) and \(f_{y}\) be continuous on \((a,b]\times [c,d],\) and suppose that

\[F(y)=\int_{a}^{b}f(x,y)\,dx\]

converges for some \(y_{0} \in [c,d]\) and

\[G(y)=\int_{a}^{b}f_{y}(x,y)\,dx\]

converges uniformly on \([c,d].\) Then \(F\) converges uniformly on \([c,d]\) and is given explicitly by

\[F(y)=F(y_{0})+\int_{y_{0}}^{y} G(t)\,dt,\quad c\le y\le d.\]

Moreover, \(F\) is continuously differentiable on \([c,d]\) and

\[\tag{A} F'(y)=G(y), \quad c \le y \le d,\]

where \(F'(c)\) and \(f_{y}(x,c)\) are derivatives from the right, and \(F'(d)\) and \(f_{y}(x,d)\) are derivatives from the left\(.\)

Let

\[F_{r}(y)=\int_{r}^{b}f(x,y)\,dx, \quad a\le r<b,\quad c \le y \le d.\]

Since \(f\) and \(f_{y}\) are continuous on \([r,b]\times [c,d]\), Theorem 1 implies that

\[F_{r}'(y)=\int_{r}^{b}f_{y}(x,y)\,dx, \quad c \le y \le d.\]

Therefore

\[\begin{aligned} F_{r}(y)&=&F_{r}(y_{0})+\int_{y_{0}}^{y}\left( \int_{r}^{b}f_{y}(x,t)\,dx\right)\,dt\\ &=&F(y_{0})+\int_{y_{0}}^{y}G(t)\,dt \\&&+(F_{r}(y_{0})-F(y_{0})) -\int_{y_{0}}^{y}\left(\int_{a}^{r}f_{y}(x,t)\,dx\right)\,dt, \quad c \le y \le d.\end{aligned}\]

Therefore,

\[\tag{B} \left|F_{r}(y)-F(y_{0})-\int_{y_{0}}^{y}G(t)\,dt\right| \le |F_{r}(y_{0})-F(y_{0})| +\left|\int_{y_{0}}^{y} \int_{a}^{r}f_{y}(x,t)\,dx\right|\,dt.\]

Now suppose \(\epsilon>0\). Since we have assumed that \(\lim_{r\to a+}F_{r}(y_{0})=F(y_{0})\) exists, there is an \(r_{0}\) in \((a,b)\) such that

\[|F_{r}(y_{0})-F(y_{0})|<\epsilon,\quad a<r< r_{0}.\]

Since we have assumed that \(G(y)\) converges for \(y\in[c,d]\), there is an \(r_{1} \in (a,b]\) such that

\[\left|\int_{a}^{b}f_{y}(x,t)\,dx\right|<\epsilon, \quad t\in[c,d], \quad a<r\le r_{1}.\]

Therefore, (B) yields

\[\left|F_{r}(y)-F(y_{0})-\int_{y_{0}}^{y}G(t)\,dt\right|< \epsilon(1+|y-y_{0}|) \le \epsilon(1+d-c)\]

if \(a<r<\min(r_{0},r_{1})\) and \(t\in[c,d]\). Therefore \(F(y)\) converges uniformly on \([c,d]\) and

\[F(y)=F(y_{0})+\int_{y_{0}}^{y}G(t)\,dt, \quad c \le y \le d.\]

Since \(G\) is continuous on \([c,d]\) by Theorem 10, (A) follows from differentiating this (Theorem 3.3.11, p. 141).

16. Since

\[|f(x)\cos xy|\le |f(x)|,\quad |f(x)\sin xy|\le |f(x)|,\text{\quad and\quad} \int_{-\infty}^{\infty} |f(x)|\,dx<\infty,\]

Theorems 6 and 7 imply that \(\int_{-\infty}^{\infty}f(x)\cos xy\,dx\) and \(\int_{-\infty}^{\infty}f(x) \sin xy\,dx\) converge uniformly on \((-\infty,\infty)\), so Theorem 10 implies that \(C(y)\) and \(S(y)\) are continuous on \((-\infty,\infty)\).

If \(y\ne0\), integrating by parts yields

\[\begin{aligned} C(y)&=&f(x)\frac{\sin xy}{y}\biggr|_{a}^{\infty}-\frac{1}{y} \int_{a}^{\infty}f'(x)\sin xy \,dx\\ &=&-f(a)\frac{\sin ay}{y} -\frac{1}{y}\int_{a}^{\infty}f'(x)\sin xy \,dx\end{aligned}\]

and

\[\begin{aligned} S(y)&=&-f(x)\frac{\cos xy}{y}\biggr|_{a}^{\infty}+\frac{1}{y} \int_{a}^{\infty}f'(x)\cos xy \,dx \\ &=&f(a)\frac{\cos ay}{y}+ \frac{1}{y}\int_{a}^{\infty}f'(x)\cos xy \,dx,\end{aligned}\]

since \(\lim_{x\to\infty} f(x)=0\). From Exercise [exer:17] with \(f\) replaced by \(f'\), \(\int_{1}^{\infty}f'(x)\cos xy \,dx\) and \(\int_{1}^{\infty}f'(x)\cos xy\,dx\) are continuous on \((-\infty,\infty)\). Therefore \(C(y)\) and \(S(y)\) are continuous on \((-\infty,0)\cup(0,\infty)\).

To see that \(C\) and \(S\) are not necessarily continuous at \(y=0\), let \(a=1\) and \(f(x)=1/x\), so

\[\lim_{x\to\infty}f(x)=0\text{\quad and\quad} \int_{1}^{\infty}|f'(x)|=\int_{1}^{\infty}\frac{dx}{x^{2}}=1.\]

Then

\[C(y)=\lim_{r\to\infty}\int_{1}^{r}\frac{\cos xy}{x}\,dx \text{\quad and\quad} S(y)=\lim_{r\to\infty}\int_{1}^{r}\frac{\sin xy}{x}\,dx,\quad y\ne0.\]

If \(y>0\) make the change of variable \(u=xy\) to see that

\[C(y)=\lim_{r\to\infty}\int_{y}^{ry}\frac{\cos u}{u}\,du= \int_{y}^{\infty}\frac{\cos u}{u}\,du\]

and

\[S(y)=\lim_{r\to\infty}\int_{y}^{ry}\frac{\sin u}{u}\,du. S(y)=\int_{y}^{\infty}\frac{\sin u}{u}\,du.\]

Therefore \(\lim_{y\to 0+}C(y)=\infty\), so \(C\) is not continuous at \(y=0\). Since \(S(0)=0\) and \(\lim_{y\to 0+}S(y)= \displaystyle{\int_{0}^{\infty}\frac{\sin u}{u}\,du}\ne 0\), \(S\) is not continuous at \(y=0\).

The integral diverges if \(y=0\). If \(y\ne0\) substitute \(u=|y|x\) to obtain

\[F(y)=\int_{0}^{\infty}\frac{dx}{1+x^{2}y^{2}}= \frac{1}{|y|}\int_{0}^{\infty}\frac{du}{1+u^{2}} =\frac{1}{|y|}\tan^{-1}u\biggr|_{0}^{\infty}=\frac{\pi}{2|y|}, \tag{A}\]

so \(F(y)\) converges for all \(y\ne0\). To test for uniform convergence, suppose \(|y|>0\) and \(0<r<r_{1}\). Then

\[\int_{r}^{r_{1}}\frac{dx}{1+x^{2}y^{2}} =\frac{1}{|y|}\int_{r|y|}^{r_{1}|y|} \frac{du}{1+u^{2}} <\frac{1}{\rho}\int_{r\rho}^{\infty}\frac{du}{1+u^{2}}\]

if \(|y|\ge \rho\). If \(\epsilon>0\) there is an \(\alpha>0\) such that \(\displaystyle{\frac{1}{\rho}\int_{\alpha}^{\infty}\frac{du}{1+u^{2}}}<\epsilon\). Therefore \(\displaystyle{\int_{r}^{r_{1}}\frac{dx}{1+x^{2}y^{2}}}<\epsilon\) if \(\alpha/\rho<r<r_{1}\). Now Theorem 4 implies that \(F(y)\) converges uniformly on \((-\infty,-\rho]\cup[\rho,\infty)\) if \(\rho>0\).

To evaluate

\[I=\displaystyle{\int_{0}^{\infty}\frac{\tan^{-1}ax-\tan^{-1}bx}{x}\,dx},\]

we note that

\[\frac{\tan^{-1}ax-\tan^{-1}bx}{x}=\int_{b}^{a}\frac{dy}{1+x^{2}y^{2}}.\]

Therefore

\[I=\int_{0}^{\infty}\,dx \int_{b}^{a}\frac{dy}{1+x^{2}y^{2}} =\int_{b}^{a}\,dy\int_{0}^{\infty}\frac{dx}{1+x^{2}y^{2}} =\frac{\pi}{2}\int_{b}^{a}\frac{dy}{y}=\frac{\pi}{2}\log\frac{a}{b},\]

where the second equality is valid because of the uniform convergence of \(F(y)\) on the closed interval with endpoints \(a\) and \(b\), and the third equality follows from (A).

\(F(y)\) is a proper integral if \(y\ge 0\) and it diverges if \(y\le -1\). If \(-1<y<0\), then

\[F(y)=\int_{0}^{1}x^{y}\,dx=\frac{x^{y+1}}{y+1}\biggr|_{0}^{1}=\frac{1}{y+1} \tag{A}\]

is convergent. Since

\[\int_{0}^{r}x^{y}\,dx=\frac{x^{y+1}}{y+1}\biggr|_{0}^{r}= \frac{r^{y+1}}{y+1}.\]

and

\[\frac{\partial}{\partial y}\left (\frac{r^{y+1}}{y+1}\right) =\frac{r^{y+1}}{y+1}\left(\log r-\frac{1}{y+1}\right)<0 \text{\quad if \quad} 0<r\le 1 \text{\quad and\quad} y>-1,\]

it follows that

\[\left|\int_{0}^{r}x^{y}\,dx\right|\le \frac{r^{\rho+1}}{\rho +1} \text{\quad if\quad} 0<r\le 1\text{\quad and\quad} -1<\rho\le y.\]

Therefore, Theorem 5 implies that \(F(y)\) converges uniformly on \([\rho,\infty)\) if \(\rho>-1\).

Now Theorem 11 implies that

\[I=\displaystyle{\int_{0}^{1}\frac{x^{a}-x^{b}}{\log x}\,dx} = \int_{0}^{1}\,dx \int_{b}^{a}x^{y}\,dy =\int_{b}^{a}\,dy\int_{0}^{1}x^{y}\,dx =\int_{b}^{a}\frac{dy}{y+1}=\log\frac{a+1}{b+1}.\]

\(\displaystyle{F(y)=\int_{0}^{\infty} e^{-yx}\cos x \,dx=\frac{y}{y^{2}+1}}\). Since

\[\left|\int_{r}^{\infty}e^{-yx} \cos x\,dx\right|\le \int_{r}^{\infty}e^{-xy}\,dx=\frac{e^{-yr}}{y},\]

Theorem 4 (or Theorem 6) implies that \(F(y)\) converges uniformly on \([\rho,\infty)\) if \(\rho>0\). Therefore, Theorem implies that if \(a\), \(b>0\) then

\[\begin{aligned} I&=&\int_{0}^{\infty}\frac{e^{-ax}-e^{-bx}}{x}\cos x\,dx =\int_{0}^{\infty}\cos x\,dx\int_{a}^{b}e^{-yx}\,dy \\ &=&\int_{a}^{b} \,dy\int_{0}^{\infty}e^{-yx}\cos x\,dx =\int_{a}^{b}\frac{y}{y^{2}+1}\,dy=\frac{1}{2}\log\frac{b^{2}+1}{a^{2}+1}.\end{aligned}\]

\(\displaystyle{F(y)=\int_{0}^{\infty} e^{-yx}\sin x \,dx=\frac{1}{y^{2}+1}}\). Since

\[\left|\int_{r}^{\infty}e^{-yx} \sin x\,dx\right|\le \int_{r}^{\infty}e^{-yx}\,dx=\frac{e^{-yr}}{y},\]

Theorem 4 (or Theorem 6) implies that \(F(y)\) converges uniformly on every \([\rho,\infty)\) if \(\rho>0\). Therefore, if \(a\), \(b>0\) then Theorem 11 implies that

\[\begin{aligned} I&=&\int_{0}^{\infty}\frac{e^{-ax}-e^{-bx}}{x}\sin x\,dx =\int_{0}^{\infty}\sin x\,dx\int_{a}^{b}e^{-yx}\,dy \\ &=&\int_{a}^{b} \,dy\int_{0}^{\infty}e^{-yx}\sin x\,dx =\int_{a}^{b}\frac{1}{y^{2}+1}\,dy=\tan^{-1}b-\tan^{-1}a.\end{aligned}\]

(e) \(\displaystyle{F(y)=\int_{0}^{\infty} e^{-x}\sin xy \,dx=\frac{y}{y^{2}+1}}\). Since

\[\left|\int_{r}^{\infty}e^{-x} \sin xy\,dx\right|\le \int_{r}^{\infty}e^{-x}\,dx=e^{-r},\]

Theorem 4 (or Theorem 6) implies that \(F(y)\) converges uniformly on \([\rho,\infty)\) if \(\rho>0\). Therefore Theorem 11 implies that

\[\begin{aligned} I&=&\int_{0}^{\infty}e^{-x}\frac{1-\cos ax}{x}\,dx =\int_{0}^{\infty}e^{-x}\,dx\int_{0}^{a}\sin xy\,dy\\ &=&\int_{0}^{a}\,dy\int_{0}^{\infty}e^{-x}\sin xy\,dx =\int_{0}^{a} \frac{y}{y^{2}+1}\,dy=\frac{1}{2}\log(1+a^{2}).\end{aligned}\]

\(\displaystyle{F(y)=\int_{0}^{\infty} e^{-x}\cos xy \,dx=\frac{1}{y^{2}+1}}\). Since

\[\left|\int_{r}^{\infty}e^{-x} \cos xy\,dx\right|\le \int_{r}^{\infty}e^{-x}\,dx=e^{-r},\]

Theorem 4 (or Theorem 6) implies that \(F(y)\) converges uniformly on \([\rho,\infty)\) if \(\rho>0\). Therefore Theorem 11 implies that

\[\begin{aligned} I&=&\int_{0}^{\infty}e^{-x}\sin ax\,dx =\int_{0}^{\infty}e^{-x}\,dx\int_{0}^{a}\cos xy\,dy\\ &=&\int_{0}^{a}\,dy\int_{0}^{\infty}e^{-x}\cos xy\,dx =\int_{0}^{a} \frac{1}{y^{2}+1}\,dy=\tan^{-1}a.\end{aligned}\]

19. (a) We start with

\[F(y)=\int_{0}^{1} x^{y}\,dx =\frac{1}{y+1}\quad y>-1. \tag{A}\]

Formally differentiating this yields

\[F^{(n)}(y)=\int_{0}^{1}(\log x)^{n} x^{y}\,dx =\frac{(-1)^{n}n!}{(y+1)^{n+1}},\quad y>-1. \tag{B}\]

To justify this we will show by induction that the improper integrals

\[I_{n}(y)=\int_{0}^{1}(\log x)^{n} x^{y}\,dx,\quad n=0,1,2, \dots\]

converge uniformly on \([\rho,\infty)\) if \(\rho>-1\). We begin with \(n=0\):.

\[\int_{0}^{r}x^{y}\,dx = \frac{x^{y+1}}{y+1}\biggr|_{r_{1}}^{r}=\frac{r^{y+1}}{y+1}\le \frac{r^{y+1}}{\rho+1},\quad -1<\rho\le y.\]

so \(I_{0}(y)=F(y)\) converges uniformly on \([\rho,\infty)\) if \(\rho>-1\). Now suppose that \(I_{n}(y)\) converges uniformly on \([\rho,\infty)\). Integrating by parts yields

\[\begin{aligned} \int_{r_{1}}^{r}(\log x)^{n+1}x^{y}\,dx&=& \frac{r^{y+1}(\log r)^{n+1}-r_{1}^{y+1}(\log r_{1})^{n+1}} {y+1}\\ &&-\frac{n+1}{y+1} \int_{r_{1}}^{r}(\log x)^{n}x^{y}\,dx, \quad -1<y<\infty.\end{aligned}\]

Letting \(r_{1}\to 0\) yields

\[\tag{C} \int_{0}^{r}(\log x)^{n+1} x^{y}\,dx =\frac{r^{y+1}(\log r)^{n+1}}{y+1} -\frac{n+1}{y+1}\int_{0}^{r}(\log x)^{n}x^{y}\,dx.\]

Since the integral on the right converges, it follows that the integral on the left converges; in fact

\[\int_{0}^{1}(\log x)^{n+1}x^{y}\,dx= -\frac{n+1}{y+1} \int_{0}^{1}(\log x)^{n}x^{y}\,dx.\]

We must still show that the integral on the left converges uniformly on \([\rho,\infty)\) if
\(\rho>-1\). To this end, note from (C) that

\[\tag{D} \left|\int_{0}^{r}(\log x)^{n+1} x^{y}\,dx\right| \le \left|\frac{r^{\rho+1}(\log r)^{n+1}}{\rho+1}\right| +\frac{n+1}{\rho+1}\left|\int_{0}^{r}(\log x)^{n}x^{y}\,dx\right|\]

if \(y\ge \rho\), Now suppose \(\epsilon>0\). Since \(\displaystyle{\lim_{r\to0+}r^{\rho+1}(\log r)^{n+1}=0}\), there is an \(r_{1}\in (0,1)\) such that

\[\left|\frac{r^{\rho+1}(\log r)^{n+1}}{\rho+1}\right| \le \frac{\epsilon}{2} \text{\quad if \quad} 0<r<r_{1}.\]

Since \(\int_{0}^{r}(\log x)^{n}x^{y}\,dx\) is uniformly convergent (by our induction assumption), there is \(r_{2}\in (0,1)\) such that

\[\frac{n+1}{\rho+1}\left|\int_{0}^{r}(\log x)^{n}x^{y}\,dx\right|\le \frac{\epsilon}{2},\quad y\ge \rho,\]

Now (D) implies that

\[\left|\int_{0}^{r}(\log x)^{n+1} x^{y}\,dx\right|\epsilon,\quad y\in [\rho,\infty),\quad 0<r<\min(r_{1},r_{2}).\]

This, Theorem 11, and an easy induction argument imply (B).

(b) Substituting \(x=u\sqrt{y}\) yields

\[F(y)=\int_{0}^{\infty}\frac{dx}{x^{2}+y}=\frac{1}{\sqrt{y}}\int_{0}^{\infty} \frac{du}{u^{2}+1} =\frac{\pi}{2\sqrt{y}},\quad y>0. \tag{A}\]

Formally differentiating this yields yields

\[\begin{aligned} \int_{0}^{\infty}\frac{dx}{(x^{2}+y)^{n+1}} &=&\frac{\pi}{2n+1}1\cdot 3\cdots(2n-1)y^{-n-1/2} =\frac{\pi}{2^{2n+1}}\frac{(2n)!}{n!}y^{-n-1/2}\\ &=&\frac{\pi}{2^{2n+1}}\binom{2n}{n}y^{-n-1/2},\quad y>0.\end{aligned}\]

Theorem 11 implies that the formal differentiation is legitimate, since, if \(y\ge 0\) and \(r>0\), then

\[\int_{r}^{\infty}\frac{dx}{(x^{2}+y)^{n+1}}\le \int_{r}^{\infty}x^{-2n-2}dx=\frac{r^{-2n-1}}{(2n-1)};\]

hence, the improper integrals \(\displaystyle{\int_{0}^{\infty}\frac{dx}{(x^{2}+y)^{n+1}}}\), \(n=0\), \(1\), \(2\), … converge uniformly on \([0,\infty)\).

Denote \(I_{n}(y)=\displaystyle{\int_{0}^{\infty}x^{2n+1}e^{-yx^{2}}\,dx}\). Then

\[I_{0}(y)=\int_{0}^{\infty}xe^{-yx^{2}}= \frac{1}{2}\int_{0}^{\infty}2xe^{-yx^{2}}\,dx =-\frac{1}{2y}e^{-yx^{2}}\biggr|_{0}^{\infty}=\frac{1}{2y}.\]

Since

\[\int_{r}^{\infty}x^{2n+1}e^{-yx^{2}} \,dx\le \int_{r}^{\infty}x^{2n+1}e^{-\rho x^{2}} \,dx\text{\quad if\quad} 0<\rho\le r,\]

if \(n\ge 0\), we can differentiate \(I_{n}\) formally with respect to \(y\in (0,\infty)\) to obtain

\[I_{n}(y)=(-1)^{n}I_{0}^{(n)}=\frac{n!}{2y^{n+1}}.\]

(d) Denote

\[\begin{aligned} I(y)&=&\int_{0}^{\infty}y^{x}\,dx =\int_{0}^{\infty}e^{x\log y}\,dx =\frac{1}{\log y}\int_{0}^{\infty}(\log y) y^{x}\,dx \\ &=&\frac{y^{x}}{\log y}\biggr|_{0}^{\infty}=-\frac{1}{\log y}\quad 0<y<1.\end{aligned}\]

Formally differentiating this yields \(I'(y)=\displaystyle{\int_{0}^{\infty}xy^{x-1}\,dx}\). There are two improper integrals here: \(J_{1}(y)=\displaystyle{\int_{0}^{1}xy^{x-1}\,dx}\) and \(J_{2}(y)=\displaystyle{\int_{1}^{\infty}xy^{x-1}\,dx}\). If \(r<1\) then

\[\int_{0}^{r}xy^{x-1}\,dx=\frac{1}{y}\int_{0}^{r}xy^{x}\,dx \le \frac{1}{y}\int_{0}^{r}x\,dx=\frac{r^{2}}{2y}\le \frac{r^{2}}{2\rho_{1}}, \quad 0<\rho_{1}\le y\le 1.\]

Therefore \(J_{1}(y)\) converges uniformly on \([\rho_{1},1]\). If \(x>r>1\) and \(\rho_{2}<1\) then

\[\int_{r}^{\infty}xy^{x-1}\,dx<\int_{r}^{\infty}x\rho_{2}^{x-1}\,dx =\frac{1}{\rho_{2}}\int_{r}^{\infty}x\rho_{2}^{x}\,dx,\]

Since

\[\lim_{x\to\infty}\frac{1}{\rho_{2}}\int_{r}^{\infty}x\rho_{2}^{x}\,dx =0\]

Theorem 7 implies that \(J_{2}(y)\) converges uniformly on \([0,\rho_{2}]\). Therefore, if \(0<\rho_{1}<\rho_{2}<1\) then \(\displaystyle{\int_{0}^{\infty}xy^{x-1}}\) converges uniformly on \([\rho_{1},\rho_{2}]\). Now Theorem 11 implies that

\[\tag{A} I'(y)=\int_{0}^{\infty}xy^{x-1}\,dx,\quad 0<y<1.\]

However, since \(I(y)=-\displaystyle{\frac{1}{\log y}}\), we know that \(I'(y)=\displaystyle{\frac{1}{y(\log y)^{2}}}\). This and (A) imply that \(\displaystyle{\int_{0}^{\infty}xy^{x}\,dx}=\displaystyle{\frac{1}{(\log x)^{2}}}\).

20. Here \(F(y)=\displaystyle{\int_{0}^{\infty} e^{-x^{2}}\cos 2xy\,dx}\), so Theorem 11 implies that

\[F'(y)=-2\int_{0}^{\infty}xe^{-x^{2}}\sin 2xy\,dx, \tag{A}\]

since the integral on the right converges uniformly on \((-\infty,\infty)\), by Theorem 6.

Integration by parts yields

\[\begin{aligned} F(y)&=& =\frac{1}{2y}\int_{0}^{\infty}e^{-x^{2}}(2y\cos 2xy)\,dx\\ &=&\frac{1}{2y}\left(e^{-x^{2}}\sin 2xy\,dx\biggr|_{0}^{\infty} +2\int_{0}^{\infty}xe^{-x^{2}}\sin 2xy\,dx\right) \\ &=&\frac{1}{y} \int_{0}^{\infty}xe^{-x^{2}}\sin 2xy\,dx =-\frac{1}{2y} F'(y).\end{aligned}\]

From this and (A), \(F'(y)+2yF(y)=0\), so \(\displaystyle{\frac{F'(y)}{F(y)}}=-2y\), \(\log F(y)=-y^{2}+\log F(0)\), and \(F(y)=F(0)e^{-y^{2}}\). Since \(F(0)=\displaystyle{\int_{0}^{\infty}e^{-x^{2}}\,dx}=\displaystyle{\frac{\sqrt{\pi}}{2}}\) (Example 12), \(F(y)=\displaystyle{\frac{\sqrt{\pi}}{2}}e^{-y^{2}}\).

Here \(F(y)=\displaystyle{\int_{0}^{\infty} e^{-x^{2}}\sin 2xy\,dx}\), so Theorem 11 implies that

\[F'(y)=2\int_{0}^{\infty}xe^{-x^{2}}\cos 2xy\,dx, \tag{A}\]

since the integral on the right converges uniformly on \((-\infty,\infty)\). Integrating this by parts yields

\[\begin{aligned} F'(y) &=&-e^{-x^{2}}\cos 2xy\biggr|_{0}^{\infty}- 2y\int_{0}^{\infty} e^{-x^{2}}\sin 2xy\,dx \\ &=&1-2y F(y),\end{aligned}\]

so \(F'(y)+2yF(y)=1\), \(e^{y^{2}}F'(y)+2e^{y^{2}}yF(y)=e^{y^{2}}\), and \(\displaystyle{\left(e^{y^{2}}F(y)\right)'=e^{y^{2}}}\). Therefore, since \(F(0)=0\), \(F(y)=\displaystyle{e^{-y^{2}}\int_{0}^{y}e^{u^{2}}\,du}\).

22. Theorems 6 and 11 imply that \(S\) and \(C\) are \(n\) times continuously differentiable on \((-\infty,\infty)\) if \(\displaystyle{\int_{-\infty}^{\infty}|x^{n}f(x)|\,dx<\infty}\).

23. We will show first that

\[C_{k}(y)=\int_{a}^{\infty} x^{k}f(x) \cos xy \,dx \text{\: and\:} S_{k}(y)=\int_{a}^{\infty}x^{k}f(x)\sin xy\,dx,\: 0\le k\le n,\]

converge uniformly on \(U_{\rho}=(-\infty,-\rho]\cup[\rho,\infty)\) if \(\rho>0\). Note that if \(\lim_{x\to\infty} x^{n}f(x)=0\), then \(\lim_{x\to\infty} x^{k}f(x)=0\), \(k=0\), \(1\), \(2\),…\(n\). If \(0\le k\le n\), then

\[\tag{A} \int_{r}^{r_{1}}x^{k}f(x)\cos xy\,dx= \frac{1}{y}\left[x^{k}f(x)\sin xy\biggr|_{r}^{r_{1}}- \int_{r}^{r_{1}}(x^{k}f(x))'\sin xy\,dx\right].\]

Henceforth \(k\) is fixed. Our assumptions imply that if \(\rho>0\) and \(\epsilon>0\) then there is an \(r_{0}\in [a,\infty)\) such that

\[\int_{r_{0}}^{\infty}|(x^{k}f(x))'|\,dx<\rho\epsilon \text{\quad and \quad} |x^{k}f(x)|<\rho\epsilon,\quad x\ge r_{0}.\]

Therefore (A) implies that

\[\left|\int_{r}^{r_{1}}x^{k}f(x)\cos xy\,dx\right|<3\epsilon,\quad r_{0}\le r<r_{1},\: y\in (-\infty,-\rho]\cup[\rho,\infty).\]

Now Theorem 4 implies that \(C_{0}\), \(C_{1}\),…, \(C_{k}\) converge uniformly on \((-\infty,-\rho]\cup[\rho,\infty)\). Since every \(y\ne0\) is in such an interval, Theorem 11 now implies that that if \(y\ne 0\) then

\[C^{(k)}(y)=\int_{a}^{\infty}x^{k}f(x)\sin xy\,dx,\quad 0\le k\le n.\]

A similar argument applies to \(S\), \(S'\),…\(S^{(n)}\).

Let \(I(y;r,r_{1})=\displaystyle{\int_{r}^{r_{1}}\frac{1}{x}\sin\frac{y}{x}\,dx}\). Assume for the moment that \(y\ge 0\). Substituting \(u=y/x\) yields

\[I(y;r,r_{1})=\int_{y/r_{1}}^{y/r}\left(\frac{u}{y}\right)\sin u \left(-\frac{y}{u^{2}}\right)\,du = \int_{y/r_{1}}^{y/r}\frac{\sin u}{u}\,du.\]

Therefore, since \(\displaystyle{\left|\frac{\sin u}{u}\right|}\le 1\) for all \(u\), \(|I(y;r,r_{1})|\le y/r\),\(1\le r\le r_{1}\). In fact, since \(I(-y;r,r_{1})=-I(y;r,r_{1})\), we can write \(|I(y;r,r_{1})\le |y|/r\), \(1\le r\le r_{1}\). Therefore, Theorem 4 implies that \(\displaystyle{\int_{1}^{\infty}\frac{1}{x}\sin\frac{y}{x}\,dx}\) converges uniformly on every finite interval.

Now denote \(F_{r}(y)=\displaystyle{\int_{1}^{r}\cos\frac{y}{x}\,dx}\). substituting \(u=y/x\) yields \(F_{r}(y)=y\displaystyle{\int_{y/r}^{y}\frac{\cos u}{u^{2}}\,du}\), so \(\lim_{r\to\infty}F_{r}(y)=\infty\) for all \(y\ge 0\). Since \(F_{r}(-y)=F_{r}(y)\), it follows that \(\lim_{r\to\infty}F_{r}(y)=\infty\) for all \(y\), so the answer to the question is “no.”

Let \(P_{n}\) be the induction assumption

\[F^{(n)}(s)=(-1)^{n}\int_{0}^{\infty}e^{-sx}x^{n}f(x)\,dx,\quad s>s_{0},\]

which is true by the definition of \(F\) for \(n=0\). If \(P_{n}\) is true, then Theorems 11 and 13 imply that

\[\begin{aligned} F^{(n+1)}(s)&=&(-1)^{n}\frac{d}{ds} \int_{0}^{\infty}e^{-sx}x^{n}f(x)\,dx=(-1)^{n} \int_{0}^{\infty}\frac{d}{ds}\left(e^{-sx}x^{n}f(x)\right)\,dx\\ &=&(-1)^{n+1}\int_{0}^{\infty}e^{-sx}x^{n+1}f(x)\,dx,\end{aligned}\]

so \(P_{n}\) implies \(P_{n+1}\), which completes the induction proof.

Let \(G(x)=\displaystyle{\int_{0}^{x}e^{-s_{0}t}f(t)\,dt}\). If \(s>s_{0}\) then

\[\tag{A} F(s)=\int_{0}^{\infty}e^{-sx}f(x)\,dx =\int_{0}^{\infty}e^{-(s-s_{0})x}G'(x)\,dx =(s-s_{0})\int_{0}^{\infty}e^{-(s-s_{0})x}G(x)\,dx\]

(integration by parts). Since \(\displaystyle{(s-s_{0})\int_{0}^{\infty}e^{-(s-s_{0})x}\,dx=1}\), (A) implies that

\[\tag{B} F(s)-F(s_{0})=(s-s_{0})\int_{0}^{\infty}e^{-(s-s_{0})x}(G(x)-F(s_{0}))\,dx.\]

Now suppose \(\epsilon>0\). Since \(F(s_{0})=\displaystyle{\int_{0}^{\infty}e^{-s_{0}t} f(t)\,dt}=\lim_{t\to\infty}G(x)\), there is an \(r\) such that \(|G(x)-F(s_{0})|<\epsilon\) if \(x\ge r\); hence, from (B), then

\[\begin{aligned} |F(s)-F(s_{0})|&\le& (s-s_{0})\int_{0}^{r}e^{-(s-s_{0})x} |G(x)-F(s_{0})|+\epsilon(s-s_{0})\int_{r}^{\infty} e^{-(s-s_{0})x}\,dx\\ &<& (s-s_{0})\int_{0}^{r}e^{-(s-s_{0})x} |G(x)-F(s_{0})|+\epsilon.\end{aligned}\]

Since \(r\) is fixed, we can let \(s\to s_{0}^{+}\) to conclude that \(\limsup_{s\to s_{0}+}|F(s)-F(s_{0})|\le \epsilon\), which implies that \(\lim_{s\to S_{0}+}F(s)=F(s_{0})\).

If \(s\ge s_{1}>s_{0}\) then

\[|e^{-sx}f(x)|= |e^{-(s-s_{0})x}e^{s_{0}x}f(x)|\le M e^{-(s-s_{0})x} \le M e^{-(s_{1}-s_{0})x}.\]

Since

\[\int_{0}^{\infty}Me^{-(s_{1}-s_{0})x}\,dx=\frac{M}{s_{1}-s_{0}}<\infty,\]

Theorem 6 implies the stated conclusion.

In Theorem 13 we assumed only that \(\int_{0}^{x}e^{-s_{0}u}f(u)\,du\) is bounded; here we are assuming that \(\int_{0}^{\infty}e^{-s_{0}u}f(u)\,du\) is convergent.

Let

\[G(x)=\int_{x}^{\infty}e^{-s_{0}t}f(t)\,dt \text{\quad and\quad} H(x)=\sup\left\{|G(t)|\, \big|\, t\ge x\right\}.\]

Then

\[\tag{A} |G(x)|\le H(x)\text{\quad and \quad} \lim_{x\to\infty}G(x)=\lim_{x\to\infty}H(x)=0,\]

since \(\int_{0}^{\infty}e^{-s_{0}x}f(x)\,dx\) converges. Since \(f\) is continuous on \([0,\infty)\), \(G'(x)=-e^{-s_{0}x}f(x)\). Integration by parts yields

\[\begin{aligned} \int_{r}^{\infty}e^{-sx}f(x)\,dx&=& \int_{r}^{\infty}e^{-(s-s_{0})x}(e^{-s_{0}x}f(x))\,dx =-\int_{0}^{\infty}e^{-(s-s_{0})x}G'(x)\,dx\\ &=&-e^{-(s-s_{0})x}G(x)\biggr|_{r}^{\infty} +(s-s_{0})\int_{r}^{\infty}e^{-(s-s_{0})x}G(x)\,dx\\ &=&e^{-(s-s_{0})r}G(r)+(s-s_{0})\int_{r}^{\infty}e^{-(s-s_{0})x}G(x)\,dx,\quad s\ge s_{0}.\end{aligned}\]

Therefore

\[\begin{aligned} \left|\int_{r}^{\infty}e^{-sx}f(x)\,dx\right|&\le& |G(r)|e^{-(s-s_{0})r}+H(r)(s-s_{0})\int_{r}^{\infty}e^{-(s-s_{0})x}\,dx\\ &=&(G(r)+H(r))e^{-(s-s_{0})r}\le 2H(r)e^{-(s-s_{0})}, \quad s\ge s_{0},\end{aligned}\]

so (A) implies that \(F(s)\) converges uniformly on \([s_{0},\infty)\).

From Theorem 13, \(F(s)=\displaystyle{\int_{0}^{\infty}e^{-sx}f(x)\, dx}\) converges for all \(s>s_{0}\). Denote \(G(x)=\displaystyle{\int_{0}^{x}e^{-s_{0}t}f(t)\,dt}\), \(x\ge 0\). Then

\[\begin{aligned} F(s)&=&\int_{0}^{\infty}e^{-(s-s_{0})x}e^{-s_{0}x}f(x)\,dx= \int_{0}^{\infty}e^{-(s-s_{0})x}G'(x)\,dx \\ &=&(s-s_{0})\int_{0}^{\infty}e^{-(s-s_{0})x}G(x)\,dx\end{aligned}\]

(integration by parts). Since \(\displaystyle{(s-s_{0})\int_{0}^{\infty}e^{-(s-s_{0})x}\,dx}=1\),

\[F(s)-F(s_{0})=\int_{0}^{\infty}e^{-(s-s_{0})x}(G(x)-F(s_{0}))\,dx\]

If \(\epsilon>0\) there is an \(R\) such that \(|G(x)-F(s_{0})|<\epsilon\) if \(x\ge R\). Therefore, if \(s>s_{0}\) then

\[\begin{aligned} |F(s)-F(s_{0})|&<& (s-s_{0})\int_{0}^{R}e^{-(s-s_{0})x}|G(x)-F(s_{0})|\,dx+\epsilon\\ &<&(s-s_{0})\int_{0}^{R}|G(x)-F(s_{0})|\,dx+\epsilon.\end{aligned}\]

Hence \(\limsup_{s\to s_{0}+}|F(s)-F(s_{0})|\le \epsilon\). Since \(\epsilon\) is arbitrary, this implies that
\(\lim_{s\to s_{0}+}|F(s)-F(s_{0})|=0\).

The assumptions of Exercise 28 imply that \(\int_{r}^{\infty}e^{-s_{0}x}f(x)\,dx\) converges for every \(r>0\). Since

\[\int_{r}^{\infty}e^{-s_{0}x}f(x)\,dx=\int_{0}^{\infty}e^{-s(r+x)}f(x+r)\,dx =e^{-sr}\int_{0}^{\infty}e^{-sx}f(x+r)\,dx,\]

we can apply the result of Exercise 30 with \(f(x)\) replaced by \(f(x+r)\), to conclude that

\[\begin{aligned} \lim_{s\to s_{0}+}\int_{r}^{\infty}e^{-sx}f(x)\,dx&=& e^{-s_{0}r}\int_{0}^{\infty}e^{-s_{0}x}f(x+r)\,dx\\ &=&\int_{0}^{\infty}e^{-s_{0}(x+r)}f(x+r)\,dx\\ &=&\int_{r}^{\infty}e^{-s_{0}x}f(x)\,dx.\end{aligned}\]

If \(G(x)=\displaystyle{\int_{0}^{x}e^{-s_{0}t}f(t)\,dt}\), then \(|G(x)|\le M\) on \([0,\infty)\) for some \(M\). If \(\epsilon>0\), there is an \(r>0\) such that

\[\tag{A} \int_{0}^{r}e^{-s_{0}x}|f(x)|\,dx <\epsilon.\]

If \(s>s_{0}\), then

\[\begin{aligned} \int_{r}^{\infty}e^{-sx}f(x)\,dx&=&\int_{r}^{\infty}e^{-(s-s_{0})x}G'(x)\,dx \\ &=&e^{-(s-s_{0})x}G(x)\biggr|_{r}^{\infty} +(s-s_{0})\int_{r}^{\infty}G(x)e^{-(s-s_{0})x}\,dx\\ &=&-e^{-(s-s_{0})r}G(r) +(s-s_{0})\int_{r}^{\infty}G(x)e^{-(s-s_{0})x}\,dx.\end{aligned}\]

Therefore, since \(|G(x)|\le M\),

\[\begin{aligned} \left|\int_{r}^{\infty}e^{-sx}f(x)\,dx\right| &\le&Me^{-(s-s_{0})r}+M(s-s_{0})\int_{r}^{\infty}e^{-(s-s_{0})x}\,dx\\ &=&M\left(e^{-(s-s_{0})r}-e^{-(s-s_{0})x}\biggr|_{r}^{\infty}\right) =2Me^{-(s-s_{0})r}.\end{aligned}\]

This and (A) imply that

\[\left|\int_{0}^{\infty}e^{-sx}f(x)\,dx\right|\le \epsilon+2Me^{-(s-s_{0})r}.\]

Therefore,

\[\limsup_{s\to\infty} \left|\int_{0}^{\infty}e^{-sx}f(x)\,dx\right|\le \epsilon.\]

Since \(\epsilon\) is arbitrary, this implies that

\[\limsup_{s\to\infty}\int_{0}^{\infty}e^{-sx}f(x)\,dx=0,\]

From Exercise 18(d), \(\displaystyle{\int_{0}^{\infty}\frac{e^{-ax}-e^{-bx}}{x}\sin x\,dx} =\tan^{-1}b-\tan^{-1}a.\) From Exercise 30, letting \(b\to\infty\) yields

\[\int_{0}^{\infty}e^{-ax}\frac{\sin x}{x}\,dx= \frac{\pi}{2}-\tan^{-1}a, \text{\quad so \bf{(b)}\quad} \int_{0}^{\infty}\frac{\sin x}{x}\,dx=\frac{\pi}{2}.\]

Integrating by parts yields

\[\tag{A} \int_{0}^{r}e^{-sx}f'(x)\,dt =e^{-sr}f(r)-f(0) +\int_{0}^{r}se^{-sx}f(x)\,dx.\]

Suppose \(s\ge s_{1}>s_{0}\). Since \(|f(x)|\le Me^{s_{0}x}\), \(e^{-sr}|f(r)|\le Me^{-(s_{1}-s_{0})r}\). Therefore \(e^{-sr}f(r)=0\) converges uniformly to zero on \([s_{1},\infty)\). Since

\[\begin{aligned} \left|\int_{r}^{\infty}se^{-sx}f(x)\,dx \right|&\le& M|s|\int_{r}^{\infty}e^{-(s-s_{0})x}\,dx\le \frac{M|s|e^{-(s_{1}-s_{0})r}}{s-s_{0}}\\ &\le&\frac{M(s-s_{0}+|s_{0}|)e^{-(s_{1}-s_{0})r}}{s-s_{0}} \le M\left(1+\frac{|s_{0}|}{s_{1}-s_{0}}\right)e^{-(s_{1}-s_{0})r},\end{aligned}\]

it follows that \(\displaystyle{\int_{r}^{\infty}se^{-sx}f(x)\,dx}\) converges to zero uniformly on \([s_{1},\infty)\). Since this implies that \(\displaystyle{\int_{0}^{r}se^{-sx}f(x)\,dx}\) converges uniformly on \([s_{1},\infty)\), (A) implies that \(G(s)\) converges uniformly on \([s_{1},\infty)\).

(b) In this case let \(f'(x)=xe^{x^{2}}\sin e^{x^{2}}\), so \(f(x)=-\displaystyle{\frac{1}{2}}\cos e^{x^{2}}\). Since \(|\cos e^{x^{2}}|\le 1\) for all \(x\), the hypotheses stated in (a) hold with \(s_{0}=0\). Therefore \(G(s)\) converges uniformly on \([\rho,\infty)\) if \(\rho>0\).

33. We will first show that \(\displaystyle{\int_{0}^{\infty}e^{-s_{0}x}\frac{f(x)}{x}\,dx}\) converges. Denote \(G(x)=\displaystyle{\int_{0}^{x}e^{-s_{0}t}f(t)\,dt}\). Since \(F(s_{0})\) is convergent; say \(|G(x)|\le M\), \(0\le x<\infty\). If \(0<r<r_{1}\) then

\[\int_{r}^{r_{1}}e^{-s_{0}x}\frac{f(x)}{x}\,dx= \int_{r}^{r_{1}}\frac{G'(x)}{x}\,dx =\frac{G(r)}{r}-\frac{G(r_{1})}{r_{1}}-\int_{r}^{r_{1}}\frac{G(x)}{x^{2}}\,dx.\]

Therefore

\[\left|\int_{r}^{r_{1}}e^{-s_{0}x}\frac{f(x)}{x}\,dx\right|\le \frac{3M}{\rho},\quad \rho<r<r_{1}\]

so Theorem 2 implies that \(H(s)=\displaystyle{\int_{0}^{\infty}e^{-st}\frac{f(x)}{x}\,dx}\) converge when \(s=s_{0}\). Therefore Exercise 27 implies that it converges uniformly on \([s_{0},\infty)\), Therefore Theorem 10 implies that

\[\begin{aligned} \int_{s_{0}}^{s}F(u)\,du &=&\int_{s_{0}}^{s}\left(\int_{0}^{\infty} e^{-ux}f(x)\,dx\right)\,du =\int_{0}^{\infty}\left(\int_{s_{0}}^{s}e^{-ux}\,du\right)f(x)\,dx\\ &=&\int_{0}^{\infty}\left(e^{-s_{0}x}-e^{-sx}\right)\frac{f(x)}{x}\,dx\end{aligned}\]

From Exercise 30 (with \(f(x)\) replaced by \(f(x)/x\)), \(\displaystyle{\lim_{s\to\infty}\int_{0}^{\infty}e^{-sx}\frac{f(x)}{x}\,dx}=0\), which implies the stated conclusion.