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.