question over a integration changes order and hard to compute

Question

I am trying to integrate the following function: $$f(x,y)=xy\times e^{-\frac{x^2y^2}{2}},x\in(1,2), y\in(0,\infty)$$ To do the integration: $$\int_1^2 \int_0^\infty f(x,y)dydx=\int_1^2\frac{1}{x}dx=\ln 2$$

However, on the other hand, if we do the integration over x first: $$ \int_0^\infty \int_1^2 f(x,y) dxdy=\int_0^\infty \frac{1}{y}(-e^{-2y^2}+e^{-0.5y^2})dy$$ This integration is hard to calculate, since each separate part is divergent, anyone can give a solution to this integration?

Substitute $t = y^2$ and then use Frullani. But you know the easy way already, so why bother? — Daniel Fischer, Oct 10 '15 at 20:06
Perform the natural substitution and it falls pretty quickly. — davidlowryduda, Oct 10 '15 at 20:07

score 0 · Accepted Answer · edited Apr 13 '17 at 12:20

0

As you have seen it is sometimes better to integrate with respect to one variable first.

The integral $$ \int_0^\infty \frac{1}{y}(-e^{-2y^2}+e^{-0.5y^2})\,dy $$ is a Frullani integral, $$ \int_0^{+\infty}\frac{f(ay)-f(by)}{y}\,dy=f(0)\log(b/a). $$

In your case $f(y)=e^{-y^2}$, $a=1/\sqrt{2}$, and $b=\sqrt{2}$.

edited Apr 13 '17 at 12:20

Community

1

answered Oct 10 '15 at 20:12

mickep

19,962

question over a integration changes order and hard to compute

1 Answers1