# Tricky Improper Integral.

• November 21st 2010, 03:08 PM
minusb
Tricky Improper Integral.
Hi, I'm trying to work through perhaps someone could advise me on this...

I'm trying to show that $\displaystyle \int_{0}^{1}(ln(1/s))^{x-1}ds$ is the same function as $\displaystyle \int_{0}^{inf}x^3e^{-x}dx$ and explain why this i

I am told that $\displaystyle \int_{0}^{inf}x^3e^{-x}dx$converges to (n-1)!

This is a little out of my league.

Here's what I'm thinking that I do a u substitution with the ln(1/s). Because I don't have an x on the bottom I don't have to use $\displaystyle \int_{a}^{1}(ln(1/s))^{x-1}ds$ right?

For the second function I use integration by parts?

Any clue as to why it's well defined at 2?

What steps would you advise me to take?
• November 21st 2010, 04:17 PM
tonio
Quote:

Originally Posted by minusb
Hi, I'm trying to work through perhaps someone could advise me on this...

I'm trying to show that $\displaystyle \int_{0}^{1}(ln(1/s))^{x-1}ds$ is the same function as $\displaystyle \int_{0}^{inf}x^3e^{-x}dx$ and explain why this i

I am told that $\displaystyle \int_{0}^{inf}x^3e^{-x}dx$converges to (n-1)!

This is a little out of my league.

Here's what I'm thinking that I do a u substitution with the ln(1/s). Because I don't have an x on the bottom I don't have to use $\displaystyle \int_{a}^{1}(ln(1/s))^{x-1}ds$ right?

For the second function I use integration by parts?

Any clue as to why it's well defined at 2?

What steps would you advise me to take?

There seem to be several typing mistakes in your question:

First, $\displaystyle \int_{0}^{1}(ln(1/s))^{x-1}ds=\displaystyle \int_{0}^{\infty}t^{x-1}e^{-t}dt=\Gamma(x)=$ the very well-known

gamma function, making the variable change $\displaystyle{\ln\frac{1}{s}=t}$.

This equals $(x-1)!$ if $x\in\mathbb{N}$ , so check why you wrote 3 instead of x-1.

Second, the second integral (not function) simply is $\Gamma(3)=2!=2$ .

Tonio
• November 21st 2010, 04:29 PM
minusb
Thanks for pointing that out Tonio.. I'll google it for a bit more info.