1. ## Improper Integrals

http://www.webassign.net/cgi-bin/sym...%20dx%20%3D%20

I know this improper integral converges, but now I don't know how to work it out and find the limit...

Any help will be appreciated!

2. Have you ever heard of the gamma distribution? It can be shown that...

$\displaystyle \int_0^{\infty} x^a e^{-bx} dx= \frac{\Gamma (a+1)}{b^{a+1}} = \frac{a!}{b^{a+1}}$

$\displaystyle \int_0^{\infty} x^1 e^{-5.9x} dx= \frac{\Gamma (2)}{(5.9)^{2}} = \frac{1!}{(5.9)^{2}}$

3. You could also just use parts:

$\displaystyle \int_0^{\infty} udv = uv|_0^{\infty} - \int_0^{\infty} vdu$

set $\displaystyle u=x$ and $\displaystyle dv= e^{-5.9x}$ to solve.

4. Originally Posted by Anonymous1
You could also just use parts:

$\displaystyle \int_0^{\infty} udv = uv|_0^{\infty} - \int_0^{\infty} vdu$

set $\displaystyle u=x$ and $\displaystyle dv= e^{-5.9x}$ to solve.
You can not do that ..
That is an improper integral, it should be evaluated in terms of limits ..
We should solve it as an indifinite integral first ..

5. Originally Posted by General
You can not do that ..
That is an improper integral, it should be evaluated in terms of limits ..
We should solve it as an indifinite integral first ..
What do you mean? I was setting the bounds of integration as-per this problem...

In any case: Use the gamma function, its easier.