1.6: Consequences of Uniform Convergence
( \newcommand{\kernel}{\mathrm{null}\,}\)
[theorem:10] If f=f(x,y) is continuous on either [a,b)×[c,d] or (a,b]×[c,d] and
F(y)=∫baf(x,y)dx
converges uniformly on [c,d], then F is continuous on [c,d]. Moreover,
∫dc(∫baf(x,y)dx)dy=∫ba(∫dcf(x,y)dy)dx.
We will assume that f is continuous on (a,b]×[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 ϵ>0, there is an r∈[a,b) such that
|∫brf(x,y)dx|<ϵ,c≤y≤d.
Therefore, if c≤y,y0≤d], then
|F(y)−F(y0)|=|∫baf(x,y)dx−∫baf(x,y0)dx|≤|∫ra[f(x,y)−f(x,y0)]dx|+|∫brf(x,y)dx|+|∫brf(x,y0)dx|,
so
|F(y)−F(y0)|≤∫ra|f(x,y)−f(x,y0)|dx+2ϵ.
Since f is uniformly continuous on the compact set [a,r]×[c,d] (Corollary 5.2.14, p. 314), there is a δ>0 such that
|f(x,y)−f(x,y0)|<ϵ
if (x,y) and (x,y0) are in [a,r]×[c,d] and |y−y0|<δ. This and [eq:25] imply that
|F(y)−F(y0)|<(r−a)ϵ+2ϵ<(b−a+2)ϵ
if y and y0 are in [c,d] and |y−y0|<δ. Therefore F is continuous on [c,d], so the integral on left side of [eq:24] exists. Denote
I=∫dc(∫baf(x,y)dx)dy.
We will show that the improper integral on the right side of [eq:24] converges to I. To this end, denote
I(r)=∫ra(∫dcf(x,y)dy)dx.
Since we can reverse the order of integration of the continuous function f over the rectangle [a,r]×[c,d] (Corollary 7.2.2, p. 466),
I(r)=∫dc(∫raf(x,y)dx)dy.
From this and [eq:26],
I−I(r)=∫dc(∫brf(x,y)dx)dy.
Now suppose ϵ>0. Since ∫baf(x,y)dx converges uniformly on [c,d], there is an r0∈(a,b] such that
|∫brf(x,y)dx|<ϵ,r0<r<b,
so |I−I(r)|<(d−c)ϵ if r0<r<b. Hence,
lim
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.