1.3: Uniform convergence of improper integrals

Henceforth we deal with functions \(f=f(x,y)\) with domains \(I\times S\), where \(S\) is an interval or a union of intervals and \(I\) is of one of the following forms:

• \([a,b)\) with \(-\infty<a<b\le \infty\);
• \((a,b]\) with \(-\infty\le a<b< \infty\);
• \((a,b)\) with \(-\infty\le a\le b\le \infty\).

In all cases it is to be understood that \(f\) is locally integrable with respect to \(x\) on \(I\). When we say that the improper integral \(\int_{a}^{b}f(x,y)\,dx\) has a stated property “on S” we mean that it has the property for every \(y\in S\).

[definition:1] If the improper integral

\[\label{eq:10} \int_{a}^{b}f(x,y)\,dx=\lim_{r\to b-}\int_{a}^{r}f(x,y)\,dx\]

converges on \(S,\) it is said to converge uniformly (or be uniformly convergent) on \(S\) if\(,\) for each \(\epsilon>0,\) there is an \(r_{0} \in [a,b)\) such that

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

or\(,\) equivalently\(,\)

\[\label{eq:11} \left|\int_{r}^{b}f(x,y)\,dx\right|< \epsilon, \quad y\in S,\quad r_{0}\le r<b.\]

The crucial difference between pointwise and uniform convergence is that \(r_{0}(y)\) in [eq:8] may depend upon the particular value of \(y\), while the \(r_{0}\) in [eq:11] does not: one choice must work for all \(y\in S\). Thus, uniform convergence implies pointwise convergence, but pointwise convergence does not imply uniform convergence.

(Cauchy Criterion for Uniform Convergence I) [theorem:4] The improper integral in [eq:10] converges uniformly on \(S\) if and only if\(,\) for each \(\epsilon>0,\) there is an \(r_{0} \in [a,b)\) such that

\[\label{eq:12} \left|\int_{r}^{r_{1}}f(x,y)\,dx\right|< \epsilon, \quad y\in S,\quad r_{0}\le r,r_{1}<b.\]

Suppose \(\int_{a}^{b} f(x,y)\,dx\) converges uniformly on \(S\) and \(\epsilon>0\). From Definition [definition:1], there is an \(r_{0}\in [a,b)\) such that

\[\label{eq:13} \left|\int_{r}^{b}f(x,y)\,dx\right| <\frac{\epsilon}{2} \text{\, and\,} \left|\int_{r_{1}}^{b}f(x,y)\,dx\right| <\frac{\epsilon}{2} ,\quad y\in S, \quad r_{0}\le r,r_{1}<b.\]

Since

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

[eq:13] and the triangle inequality imply [eq:12].

For the converse, denote

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

Since [eq:12] implies that

\[\label{eq:14} |F(r,y)-F(r_{1},y)|< \epsilon,\quad y\in S, \quad r_{0}\le r, r_{1}<b,\]

Theorem [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

\[\label{eq:15} \left|\int_{r}^{b}f(x,y)\,dx\right|< \epsilon, \quad y\in S,\quad r_{0}(y)\le r< b.\]

For each \(y\in S\), choose \(r_{1}(y)\ge \max[{r_{0}(y),r_{0}}]\). (Recall [eq:14]). Then

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

so [eq:12], [eq:15], and the triangle inequality imply that

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

In practice, we don’t explicitly exhibit \(r_{0}\) for each given \(\epsilon\). It suffices to obtain estimates that clearly imply its existence.

[example:4] For the improper integral of Example [example:2],

\[\left|\int_{r}^{\infty}f(x,y)\,dx\right|= \int_{r}^{\infty} |y|e^{-|y|x}=e^{-r|y|}, \quad y\ne0.\]

If \(|y| \ge \rho\), then

\[\left|\int_{r}^{\infty}f(x,y)\,dx\right| \le e^{-r\rho},\]

so \(\int_{0}^{\infty}f(x,y)\,dx\) converges uniformly on \((-\infty,\rho]\cup[\rho,\infty)\) if \(\rho>0\); however, it does not converge uniformly on any neighborhood of \(y=0\), since, for any \(r>0\), \(e^{-r|y|}>\frac{1}{2}\) if \(|y|\) is sufficiently small.

[definition:2] If the improper integral

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

converges on \(S,\) it is said to converge uniformly (or be uniformly convergent) on \(S\) if\(,\) for each \(\epsilon>0,\) there is an \(r_{0} \in (a,b]\) such that

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

or\(,\) equivalently\(,\)

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

We leave proof of the following theorem to you (Exercise [exer:3]).

(Cauchy Criterion for Uniform Convergence II) [theorem:5] 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

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

We need one more definition, as follows.

[definition:3] Let \(f=f(x,y)\) be defined on \((a,b) \times S,\) where \(-\infty\le a<b\le \infty.\) Suppose \(f\) is locally integrable on \((a,b)\) for all \(y\in S\) and let \(c\) be an arbitrary point in \((a,b).\) Then \(\int_{a}^{b}f(x,y)\,dx\) is said to converge uniformly on \(S\) if \(\int_{a}^{c}f(x,y)\,dx\) and \(\int_{c}^{b}f(x,y)\,dx\) both converge uniformly on \(S.\)

We leave it to you (Exercise [exer:4]) to show that this definition 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 \(S\) for every \(c \in (a,b)\).

We also leave it you (Exercise [exer:5]) to 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 three Definitions [definition:1]–[definition:3].

[example:5] Consider the improper integral

\[F(y)=\int_{0}^{\infty}x^{-1/2}e^{-xy}\,dx,\]

which diverges if \(y\le 0\) (verify). Definition [definition:3] applies if \(y>0\), so we consider the improper integrals

\[F_{1}(y)=\int_{0}^{1}x^{-1/2}e^{-xy}\,dx \text{\quad and\quad} F_{2}(y)=\int_{1}^{\infty}x^{-1/2}e^{-xy}\,dx\]

separately. Moreover, we could just as well define

\[\label{eq:16} F_{1}(y)=\int_{0}^{c}x^{-1/2}e^{-xy}\,dx \text{\quad and\quad} F_{2}(y)=\int_{c}^{\infty}x^{-1/2}e^{-xy}\,dx,\]

where \(c\) is any positive number.

Definition [definition:2] applies to \(F_{1}\). If \(0<r_{1}<r\) and \(y\ge 0\), then

\[\left|\int_{r}^{r_{1}}x^{-1/2}e^{-xy}\,dx\right| < \int_{r_{1}}^{r}x^{-1/2}\,dx<2r^{1/2},\]

so \(F_{1}(y)\) converges for uniformly on \([0,\infty)\).

Definition [definition:1] applies to \(F_{2}\). Since

\[\left|\int_{r}^{r_{1}}x^{-1/2}e^{-xy}\,dx\right| < r^{-1/2} \int_{r}^{\infty}e^{-xy}\,dx = \frac{e^{-ry}}{yr^{1/2}},\]

\(F_{2}(y)\) converges uniformly on \([\rho,\infty)\) if \(\rho>0\). It does not converge uniformly on \((0,\rho)\), since the change of variable \(u=xy\) yields

\[\int_{r}^{r_{1}}x^{-1/2}e^{-xy}\,dx=y^{-1/2} \int_{ry}^{r_{1}y}u^{-1/2}e^{-u}\,du,\]

which, for any fixed \(r>0\), can be made arbitrarily large by taking \(y\) sufficiently small and \(r=1/y\). Therefore we conclude that \(F(y)\) converges uniformly on \([\rho,\infty)\) if \(\rho>0.\)

Note that that the constant \(c\) in [eq:16] plays no role in this argument.

[example:6] Suppose we take

\[\label{eq:17} \int_{0}^{\infty}\frac{\sin u}{u}\,du =\frac{\pi}{2}\]

as given (Exercise [exer:31](b)). Substituting \(u=xy\) with \(y>0\) yields

\[\label{eq:18} \int_{0}^{\infty}\frac{\sin xy}{x}\,dx=\frac{\pi}{2},\quad y>0.\]

What about uniform convergence? Since \((\sin xy)/x\) is continuous at \(x=0\), Definition [definition:1] and Theorem [theorem:4] apply here. If \(0<r<r_{1}\) and \(y>0\), then

\[\int_{r}^{r_{1}}\frac{\sin xy}{x}\,dx=-\frac{1}{y} \left(\frac{\cos xy}{x}\biggr|_{r}^{r_{1}}+ \int_{r}^{r_{1}}\frac{\cos xy}{x^{2}}\,dx\right), \text{\, so\quad} \left|\int_{r}^{r_{1}}\frac{\sin xy}{x}\,dx\right|<\frac{3}{ry}.\]

Therefore [eq:18] converges uniformly on \([\rho,\infty)\) if \(\rho>0\). On the other hand, from [eq:17], there is a \(\delta>0\) such that

\[\int_{u_{0}}^{\infty}\frac{\sin u}{u}\,du>\frac{\pi}{4}, \quad 0 \le u_{0}<\delta.\]

This and [eq:18] imply that

\[\int_{r}^{\infty}\frac{\sin xy}{x}\,dx=\int_{yr}^{\infty}\frac{\sin u}{u}\,du >\frac{\pi}{4}\]

for any \(r>0\) if \(0 <y<\delta/r\). Hence, [eq:18] does not converge uniformly on any interval \((0,\rho]\) with \(\rho>0\).