1.6: Consequences of Uniform Convergence

[theorem:10] If $f=f(x,y)$ is continuous on either $[a,b)\times [c,d]$ or $(a,b]\times [c,d]$ and

$$\label{eq:23} F(y)=\int_{a}^{b}f(x,y)\,dx$$

converges uniformly on $[c,d],$ then $F$ is continuous on $[c,d].$ Moreover$,$

$$\label{eq:24} \int_{c}^{d}\left(\int_{a}^{b}f(x,y)\,dx\right)\,dy =\int_{a}^{b}\left(\int_{c}^{d}f(x,y)\,dy\right)\,dx.$$

We will assume that $f$ is continuous on $(a,b]\times [c,d]$. You can consider the other case (Exercise [exer:14]).

We will first show that $F$ in [eq:23] is continuous on $[c,d]$. Since $F$ converges uniformly on $[c,d]$, Definition [definition:1] (specifically, [eq:11]) implies that if $\epsilon>0$, there is an $r \in [a,b)$ such that

$$\left|\int_{r}^{b}f(x,y)\,dx\right|< \epsilon, \quad c \le y \le d.$$

Therefore, if $c\le y, y_{0}\le d]$, then

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

so

$$\label{eq:25} |F(y)-F(y_{0})| \le \int_{a}^{r}|f(x,y)-f(x,y_{0})|\,dx +2\epsilon.$$

Since $f$ is uniformly continuous on the compact set $[a,r]\times [c,d]$ (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 $[a,r]\times [c,d]$ and $|y-y_{0}|<\delta$. This and [eq:25] 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 [eq:24] exists. Denote

$$\label{eq:26} I= \int_{c}^{d}\left(\int_{a}^{b}f(x,y)\,dx\right)\,dy.$$

We will show that the improper integral on the right side of [eq:24] converges to $I$. To this end, denote

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

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

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

From this and [eq:26],

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

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

$$\left|\int_{r}^{b}f(x,y)\,dx\right|<\epsilon, \quad r_{0}<r<b,$$

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

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

which completes the proof of [eq:24].

[example:10] It is straightforward to verify that

$$\int_{0}^{\infty}e^{-xy}\,dx=\frac{1}{y}, \quad y>0,$$

and the convergence is uniform on $[\rho,\infty)$ if $\rho>0$. Therefore Theorem [theorem:10] implies that if $0<y_{1}<y_{2}$, then

$$\begin{aligned} \int_{y_{1}}^{y_{2}}\frac{\,dy}{y}&=& \int_{y_{1}}^{y_{2}}\left( \int_{0}^{\infty}e^{-xy}\,dx\right)\,dy =\int_{0}^{\infty}\left(\int_{y_{1}}^{y_{2}}e^{-xy}\,dy\right)\,dy \\ &=&\int_{0}^{\infty}\frac{e^{-xy_{1}}-e^{-xy_{2}}}{x}\,dx.\end{aligned}$$

Since

$$\int_{y_{1}}^{y_{2}}\frac{dy}{y}= \log\frac{y_{2}}{y_{1}}, \quad y_{2} \ge y_{1}>0,$$

it follows that

$$\int_{0}^{\infty}\frac{e^{-xy_{1}}-e^{-xy_{2}}}{x}\,dx= \log\frac{y_{2}}{y_{1}}, \quad y_{2} \ge y_{1}>0.$$

[example:11] From Example [example:6],

$$\int_{0}^{\infty}\frac{\sin xy}{x}\,dx=\frac{\pi}{2}, \quad y>0,$$

and the convergence is uniform on $[\rho,\infty)$ if $\rho>0$. Therefore, Theorem [theorem:10] implies that if $0<y_{1}<y_{2}$, then

$$\begin{aligned} \frac{\pi}{2}(y_{2}-y_{1}) &=&\int_{y_{1}}^{y_{2}}\left(\int_{0}^{\infty}\frac{\sin xy}{x}\,dx\right)\,dy =\int_{0}^{\infty}\left(\int_{y_{1}}^{y_{2}}\frac{\sin xy}{x}\,dy\right)\,dx \nonumber\\ &=&\int_{0}^{\infty}\frac{\cos xy_{1}-\cos xy_{2}}{x^{2}} \,dx. \label{eq:27}\end{aligned}$$

The last integral converges uniformly on $(-\infty,\infty)$ (Exercise 10(h)), and is therefore continuous with respect to $y_{1}$ on $(-\infty,\infty)$, by Theorem [theorem:10]; in particular, we can let $y_{1}\to0+$ in [eq:27] and replace $y_{2}$ by $y$ to obtain

$$\int_{0}^{\infty} \frac{1-\cos xy}{x^{2}}\,dx=\frac{\pi y}{2}, \quad y \ge 0.$$

The next theorem is analogous to Theorem 4.4.20 (p. 252).

[theorem:11] Let $f$ and $f_{y}$ be continuous on either $[a,b)\times [c,d]$ or $(a,b]\times [c,d].$ Suppose that the improper integral

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

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 $[c,d]$; specifically,

$$\label{eq:28} F'(y)=G(y), \quad c \le y \le d,$$

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$.$

We will assume that $f$ and $f_{y}$ are continuous on $[a,b)\times [c,d]$. You can consider the other case (Exercise [exer:15]).

Let

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

Since $f$ and $f_{y}$ are continuous on $[a,r]\times [c,d]$, Theorem [theorem:1] implies that

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

Then

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

Therefore,

$$\begin{aligned} \left|F_{r}(y)-F(y_{0})-\int_{y_{0}}^{y}G(t)\,dt\right|& \le & |F_{r}(y_{0})-F(y_{0})|\nonumber\\ &&+\left|\int_{y_{0}}^{y} \int_{r}^{b}f_{y}(x,t)\,dx\right|\,dt. \label{eq:29}\end{aligned}$$

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

$$|F_{r}(y_{0})-F(y_{0})|<\epsilon,\quad r_{0}<r<b.$$

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

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

Therefore, [eq:29] 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 $\max(r_{0},r_{1}) \le r <b$ 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 [theorem:10], [eq:28] follows from differentiating this (Theorem 3.3.11, p. 141).

[example:12] Let

$$I(y)=\int_{0}^{\infty}e^{-yx^{2}}\,dx, \quad y>0.$$

Since

$$\int_{0}^{r}e^{-yx^{2}}\,dx=\frac{1}{\sqrt{y}} \int_{0}^{r\sqrt{y}} e^{-t^{2}}\,dt,$$

it follows that

$$I(y)=\frac{1}{\sqrt{y}}\int_{0}^{\infty}e^{-t^{2}}\,dt,$$

and the convergence is uniform on $[\rho,\infty)$ if $\rho>0$ (Exercise [exer:8](i)). To evaluate the last integral, denote $J(\rho)=\int_{0}^{\rho}e^{-t^{2}}\,dt$; then

$$J^{2}(\rho)=\left(\int_{0}^{\rho}e^{-u^{2}}\,du\right) \left(\int_{0}^{\rho}e^{-v^{2}}\,dv\right) =\int_{0}^{\rho}\int_{0}^{\rho}e^{-(u^{2}+v^{2})}\,du\,dv.$$

Transforming to polar coordinates $r=r\cos\theta$, $v=r\sin\theta$ yields

$$J^{2}(\rho)=\int_{0}^{\pi/2}\int_{0}^{\rho} re^{-r^{2}}\,dr\,d\theta =\frac{\pi(1-e^{-\rho^{2}})}{4}, \text{\quad so\quad} J(\rho)=\frac{\sqrt{\pi(1-e^{-\rho^{2}})}}{2}.$$

Therefore

$$\int_{0}^{\infty}e^{-t^{2}}\,dt=\lim_{\rho\to\infty}J(\rho)= \frac{\sqrt{\pi}}{2}\text{\quad and\quad} \int_{0}^{\infty}e^{-yx^{2}}\,dx= \frac{1}{2}\sqrt{\frac{\pi}{y}}, \quad y>0.$$

Differentiating this $n$ times with respect to $y$ yields

$$\int_{0}^{\infty}x^{2n}e^{-yx^{2}}\,dx= \frac{1\cdot3\cdots(2n-1)\sqrt{\pi}}{2^{n}y^{n+1/2}}\quad y>0,\quad n=1,2,3, \dots,$$

where Theorem [theorem:11] justifies the differentiation for every $n$, since all these integrals converge uniformly on $[\rho,\infty)$ if $\rho>0$ (Exercise [exer:8](i)).

Some advice for applying this theorem: Be sure to check first that $F(y_{0})=\int_{a}^{b}f(x,y_{0})\,dx$ converges for at least one value of $y$. If so, differentiate $\int_{a}^{b}f(x,y)\,dx$ formally to obtain $\int_{a}^{b}f_{y}(x,y)\,dx$. Then $F'(y)=\int_{a}^{b}f_{y}(x,y)\,dx$ if $y$ is in some interval on which this improper integral converges uniformly.