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=limr→b−∫raf(x,y)dx
if, for each y∈S and every ϵ>0, there is an r=r0(y) (which also depends on ϵ) such that
|F(y)−∫raf(x,y)dx|=|∫brf(x,y)dx|<ϵ,r0(y)≤y<b.
If the domain of f is (a,b]×S where −∞≤a<b<∞, we replace [eq:7] by
∫baf(x,y)dx=limr→a+∫brf(x,y)dx
and [eq:8] by
|F(y)−∫brf(x,y)dx|=|∫raf(x,y)dx|<ϵ,a<r≤r0(y).
In general, pointwise convergence of F for all y∈S does not imply that F is continuous or integrable on [c,d], and the additional assumptions that fy is continuous and ∫bafy(x,y)dx converges do not imply [eq:2].
[example:2] The function
f(x,y)=ye−|y|x
is continuous on [0,∞)×(−∞,∞) and
F(y)=∫∞0f(x,y)dx=∫∞0ye−|y|xdx
converges for all y, with
F(y)={−1y<0,−0y=0,−1y>0;
therefore, F is discontinuous at y=0.
[example:3] The function
f(x,y)=y3e−y2x
is continuous on [0,∞)×(−∞,∞). Let
F(y)=∫∞0f(x,y)dx=∫∞0y3e−y2xdx=y,−∞<y<∞.
Then
F′(y)=1,−∞<y<∞.
However,
∫∞0∂∂y(y3e−y2x)dx=∫∞0(3y2−2y4x)e−y2xdx={1,y≠0,0,y=0,
so
F′(y)≠∫∞0∂f(x,y)∂ydx\quad if\quady=0.