# Gamma Function -Substituting Variables-

• April 24th 2010, 02:48 PM
Anthonny
Gamma Function -Substituting Variables-
The expression for the Gamma function is given as:

$\Gamma(z)=\int^\infty_0t^{z-1}e^{-t}dt$

for Re(z)>0, and we change the variable $t$ into $t=ns$

for $n\in\mathbb N$.

It then says that by this, we get:

$\Gamma(z)=n^z\int^\infty_0s^{z-1}e^{-ns}dt$

Which I am not sure how they got.

Here is what I tried:
Given $t=ns$
Then I assume $dt=nds$ and $\frac{1}{n}dt=ds$

Substituting variables, I get:

$\Gamma(z)=\frac{1}{n}\int^\infty_0(ns)^{z-1}e^{-ns}ds$

$\Gamma(z)=\frac{1}{n}\int^\infty_0(n^{z-1})s^{z-1}e^{-ns}ds$

$\Gamma(z)=\frac{n^{z-1}}{n}\int^\infty_0s^{z-1}e^{-ns}ds
$

Which is not the same as what I'm suppose to get.

If someone could show how it should be done, that would be greatly appreciated.

Thank you.
• April 24th 2010, 03:06 PM
drumist
You're making a mistake when changing $dt$ to $ds$.

$dt=n \, ds$

should give you:

$\Gamma(z)=n \int_0^{\infty} (ns)^{z-1} e^{-ns} \, ds$
• April 24th 2010, 03:10 PM
Anthonny
Quote:

Originally Posted by drumist
You're making a mistake when changing $dt$ to $ds$.

$dt=n \, ds$

should give you:

$\Gamma(z)=n \int_0^{\infty} (ns)^{z-1} e^{-ns} \, ds$

Thank you!
I completely overlooked that simple mistake.
Thanks again.