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.