1.1: Introduction to Improper Functions

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

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