Loading [MathJax]/jax/element/mml/optable/BasicLatin.js
Skip to main content
Library homepage
 

Text Color

Text Size

 

Margin Size

 

Font Type

Enable Dyslexic Font
Mathematics LibreTexts

1.1: Introduction to Improper Functions

( \newcommand{\kernel}{\mathrm{null}\,}\)

In Section 7.2 (pp. 462–484) we considered functions of the form

F(y)=baf(x,y)dx,cyd.

We saw that if f is continuous on [a,b]×[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

dcF(y)dy=dc(baf(x,y)dx)dy

to evaluate it as

dcF(y)dy=ba(dcf(x,y)dy)dx

(Corollary 7.2.3, p. 466).

Here is another important property of F.

[theorem:1] If f and fy are continuous on [a,b]×[c,d], then

F(y)=baf(x,y)dx,cyd,

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,

F(y)=bafy(x,y)dx,cyd.

Here F(a) and fy(x,a) are derivatives from the right and F(b) and fy(x,b) are derivatives from the left.

If y and y+Δy are in [c,d] and Δy0, then

F(y+Δy)F(y)Δy=baf(x,y+Δy)f(x,y)Δydx.

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

f(x,y+Δy)f(x,y)=fy(x,y)Δy=fy(x,y(x))Δy+(fy(x,y(x)fy(x,y))Δy.

From this and [eq:3],

|F(y+Δy)F(y)Δybafy(x,y)dx|ba|fy(x,y(x))fy(x,y)|dx.

Now suppose ϵ>0. Since fy is uniformly continuous on the compact set [a,b]×[c,d] (Corollary 5.2.14, p. 314) and y(x) is between y and y+Δy, there is a δ>0 such that if |Δ|<δ then

|fy(x,y)fy(x,y(x))|<ϵ,(x,y)[a,b]×[c,d].

This and [eq:4] imply that

|F(y+ΔyF(y))Δybafy(x,y)dx|<ϵ(ba)

if y and y+Δy are in [c,d] and 0<|Δy|<δ. 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 fy), F is continuous on [c,d].

[example:1] Since

f(x,y)=cosxy\quad and\quadfy(x,y)=xsinxy

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

F(y)=π0cosxydx,<y<,

then

F(y)=π0xsinxydx,<y<.

(In applying Theorem [theorem:1] for a specific value of y, we take R=[0,π]×[ρ,ρ], where ρ>|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)=sinxyy|πx=0=sinπyy,y0.

Differentiating this and using [eq:6] yields

π0xsinxydx=sinπyy2πcosπyy,y0.

To verify this, use integration by parts.

We will study the continuity, differentiability, and integrability of

F(y)=baf(x,y)dx,yS,

where S is an interval or a union of intervals, and F is a convergent improper integral for each yS. If the domain of f is [a,b)×S where <a<b, we say that F is pointwise convergent on S or simply convergent on S, and write

baf(x,y)dx=lim

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.


This page titled 1.1: Introduction to Improper Functions is shared under a CC BY-NC-SA 3.0 license and was authored, remixed, and/or curated by William F. Trench via source content that was edited to the style and standards of the LibreTexts platform.

Support Center

How can we help?