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# 1.1: Introduction to Improper Functions

• • Contributed by William F. Trench
• Andrew G. Cowles Distinguished Professor Emeritus (Mathamatics) at Trinity University
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In Section 7.2 (pp. 462–484) we considered functions of the form

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

We saw that if $$f$$ is continuous on $$[a,b]\times [c,d]$$, then $$F$$ is continuous on $$[c,d]$$ (Exercise 7.2.3, p. 481) and that we can reverse the order of integration in

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

to evaluate it as

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

(Corollary 7.2.3, p. 466).

Here is another important property of $$F$$.

[theorem:1] If $$f$$ and $$f_{y}$$ are continuous on $$[a,b]\times [c,d],$$ then

$\label{eq:1} F(y)=\int_{a}^{b}f(x,y)\,dx, \quad c \le y \le d,$

is continuously differentiable on $$[c,d]$$ and $$F'(y)$$ can be obtained by differentiating [eq:1] under the integral sign with respect to $$y;$$ that is,

$\label{eq:2} F'(y)=\int_{a}^{b}f_{y}(x,y)\,dx, \quad c \le y \le d.$

Here $$F'(a)$$ and $$f_{y}(x,a)$$ are derivatives from the right and $$F'(b)$$ and $$f_{y}(x,b)$$ are derivatives from the left$$.$$

If $$y$$ and $$y+\Delta y$$ are in $$[c,d]$$ and $$\Delta y\ne0$$, then

$\label{eq:3} \frac{F(y+\Delta y)-F(y)}{\Delta y}= \int_{a}^{b}\frac{f(x,y+\Delta y)-f(x,y)}{\Delta y}\,dx.$

From the mean value theorem (Theorem 2.3.11, p. 83), if $$x\in[a,b]$$ and $$y$$, $$y+\Delta y\in[c,d]$$, there is a $$y(x)$$ between $$y$$ and $$y+\Delta y$$ such that

$f(x,y+\Delta y)-f(x,y)=f_{y}(x,y)\Delta y= f_{y}(x,y(x))\Delta y+(f_{y}(x,y(x)-f_{y}(x,y))\Delta y.$

From this and [eq:3],

$\label{eq:4} \left|\frac{F(y+\Delta y)-F(y)}{\Delta y}-\int_{a}^{b}f_{y}(x,y)\,dx\right| \le \int_{a}^{b} |f_{y}(x,y(x))-f_{y}(x,y)|\,dx.$

Now suppose $$\epsilon>0$$. Since $$f_{y}$$ is uniformly continuous on the compact set $$[a,b]\times [c,d]$$ (Corollary 5.2.14, p. 314) and $$y(x)$$ is between $$y$$ and $$y+\Delta y$$, there is a $$\delta>0$$ such that if $$|\Delta|<\delta$$ then

$|f_{y}(x,y)-f_{y}(x,y(x))|<\epsilon,\quad (x,y)\in[a,b]\times [c,d].$

This and [eq:4] imply that

$\left|\frac{F(y+\Delta y-F(y))}{\Delta y}-\int_{a}^{b}f_{y}(x,y)\,dx\right|<\epsilon(b-a)$

if $$y$$ and $$y+\Delta y$$ are in $$[c,d]$$ and $$0<|\Delta y|<\delta$$. This implies [eq:2]. Since the integral in [eq:2] is continuous on $$[c,d]$$ (Exercise 7.2.3, p. 481, with $$f$$ replaced by $$f_{y}$$), $$F'$$ is continuous on $$[c,d]$$.

[example:1] Since

$f(x,y)=\cos xy\text{\quad and\quad} f_{y}(x,y)=-x\sin xy$

are continuous for all $$(x,y)$$, Theorem [theorem:1] implies that if

$\label{eq:5} F(y)=\int_{0}^{\pi} \cos xy\,dx,\quad -\infty<y<\infty,$

then

$\label{eq:6} F'(y)=-\int_{0}^{\pi}x\sin xy\,dx,\quad -\infty<y<\infty.$

(In applying Theorem [theorem:1] for a specific value of $$y$$, we take $$R=[0,\pi]\times [-\rho,\rho]$$, where $$\rho>|y|$$.) This provides a convenient way to evaluate the integral in [eq:6]: integrating the right side of [eq:5] with respect to $$x$$ yields

$F(y)=\frac{\sin xy}{y}\bigg|_{x=0}^{\pi}=\frac{\sin\pi y}{y}, \quad y\ne0.$

Differentiating this and using [eq:6] yields

$\int_{0}^{\pi}x\sin xy\,dx =\frac{\sin \pi y}{y^{2}}- \frac{\pi\cos \pi y}{y}, \quad y\ne0.$

To verify this, use integration by parts.

We will study the continuity, differentiability, and integrability of

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

where $$S$$ is an interval or a union of intervals, and $$F$$ is a convergent improper integral for each $$y\in S$$. If the domain of $$f$$ is $$[a,b)\times S$$ where $$-\infty<a< b\le \infty$$, we say that $$F$$ is pointwise convergent on $$S$$ or simply convergent on $$S$$, and write

$\label{eq:7} \int_{a}^{b}f(x,y)\,dx=\lim_{r\to b-}\int_{a}^{r}f(x,y)\,dx$

if, for each $$y\in S$$ and every $$\epsilon>0$$, there is an $$r=r_{0}(y)$$ (which also depends on $$\epsilon$$) such that

$\label{eq:8} \left|F(y)-\int_{a}^{r}f(x,y)\,dx\right|= \left|\int_{r}^{b}f(x,y)\,dx\right|< \epsilon, \quad r_{0}(y)\le y<b.$

If the domain of $$f$$ is $$(a,b]\times S$$ where $$-\infty\le a<b<\infty$$, we replace [eq:7] by

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

and [eq:8] by

$\left|F(y)-\int_{r}^{b}f(x,y)\,dx\right|= \left|\int_{a}^{r}f(x,y)\,dx\right|< \epsilon, \quad a<r\le r_{0}(y).$

In general, pointwise convergence of $$F$$ for all $$y\in S$$ does not imply that $$F$$ is continuous or integrable on $$[c,d]$$, and the additional assumptions that $$f_{y}$$ is continuous and $$\int_{a}^{b}f_{y}(x,y)\,dx$$ converges do not imply [eq:2].

[example:2] The function

$f(x,y)=ye^{-|y|x}$

is continuous on $$[0,\infty)\times (-\infty,\infty)$$ and

$F(y)=\int_{0}^{\infty}f(x,y)\,dx =\int_{0}^{\infty}ye^{-|y|x}\,dx$

converges for all $$y$$, with

$F(y)= \begin{cases} -1& y<0,\\ \phantom{-}0&y=0,\\ \phantom{-}1&y>0;\\ \end{cases}$

therefore, $$F$$ is discontinuous at $$y=0$$.

[example:3] The function

$f(x,y)=y^{3}e^{-y^{2}x}$

is continuous on $$[0,\infty)\times (-\infty,\infty)$$. Let

$F(y)=\int_{0}^{\infty}f(x,y)\,dx= \int_{0}^{\infty}y^{3}e^{-y^{2}x}\,dx =y,\quad -\infty<y<\infty.$

Then

$F'(y)=1, \quad -\infty<y<\infty.$

However,

$\int_{0}^{\infty}\frac{\partial{}}{\partial{y}}(y^{3}e^{-y^{2}x})\,dx =\int_{0}^{\infty}(3y^{2}-2y^{4}x)e^{-y^{2}x}\,dx= \begin{cases} 1,& y\ne0,\\ 0,& y=0, \end{cases}$

so

$F'(y)\ne\int_{0}^{\infty}\frac{\partial{f(x,y)}}{\partial{y}}\,dx\text{\quad if\quad}y=0.$