Question

SKT 6:01 Theorem lI Let f(x, y) be continuous for a s x 00, c s y 00, and let the integrals J(y)dr f(x, ) and J ( dy f(x, z) be respectively, uniformly convergent on every finite interval c sy

How to prove this theorem hold?
(It is an improper integral, extension of Riemann)

Answer 1

Such as

If f(x) and g(x) are positive and integrable, and f(x) $\leq$ g(x) when x > a,

Then,  $\int_{a}^{\infty }$ f(x) dx is convergent wen $\int_{a}^{\infty }$ g(x) dx is convergent.

Let,($\mu$ test): If x^$\mu$ is bounded for x>a , then $\int_{a}^{\infty }$ f(x) dx is absolutely convergent, provided $\mu$ > 1 and particular , if
liim
x^$\mu$ f(x) exist , where $\mu$>1, then

$\int_{a}^{\infty }$ f(x) dx converges absolutely ,

Here, | $\int_{a}^{x}$f(x) $\leq$$\int_{a}^{x}$ | f(x) | dx  $\leq$ k $\int_{a}^{x}$x-$\mu$ dx , therefore this integeral convergaes where $\mu$ >1

i.e $\int_{a}^{\infty }$ f(x) dx will converge.

Another e.g

$\int_{1}^{\infty }$x^-2 is an improper integral . Some such integral can sometimes be computed by replacing infinity limits with finite values.

$\int_{1}^{y }$x^-2 dx = 1 -1/y

amd then taking the limit as y-$\infty$,

$\int_{1}^{\infty }$ x^-2 dx =  
liim
  $\int_{1}^{y }$ x^-2 dx

=

lim 1 - 1/y y-oo

= 1

Improper Integral of the form of $\int_{a}^{b}$f(x) dx

with one infinite limit and other non zero may be expressed as finite integral. If f(x) decreased at least as fast as 1/x², then let,

t $\equiv$ 1/x

dt = -dx/x²

dx = -x² dt

= -dt / t²,

and $\int_{a}^{b}$f(x) dx =    1/t² f(1/t) dt

1/b

=  $\int_{1/b}^{1/a}$ 1/t² f(1/t) dt

If the integral diverges as exponentially. then let

t $\equiv$ e^-x

dt = -e^-x dx

x = - ln t,

An Improper integral is a definite integral that has either or both limits infinite or an integrand that approaches infinity at one or more points in the range of integration. Improper integrals can not be computed using a normal Riemann integral.