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,c≤y≤d.
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,c≤y≤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,
F′(y)=∫bafy(x,y)dx,c≤y≤d.
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 Δy≠0, 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)Δy−∫bafy(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+Δy−F(y))Δy−∫bafy(x,y)dx|<ϵ(b−a)
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,y≠0.
Differentiating this and using [eq:6] yields
∫π0xsinxydx=sinπyy2−πcosπyy,y≠0.
To verify this, use integration by parts.
We will study the continuity, differentiability, and integrability of
F(y)=∫baf(x,y)dx,y∈S,
where S is an interval or a union of intervals, and F is a convergent improper integral for each y∈S. 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.