Help with integration

• Oct 7th 2009, 02:20 PM
Chloe18
Help with integration
It's the integration of

[x^(3/2)][e^(-x)] from 0 to infinity

I tried to do integration by parts, but that didn't work out.

I know what the answer is but I don't know how to get there.

Any help would be great

Thanks
• Oct 7th 2009, 02:24 PM
artvandalay11
Quote:

Originally Posted by Chloe18
It's the integration of

[x^(3/2)][e^(-x)] from 0 to infinity

I tried to do integration by parts, but that didn't work out.

I know what the answer is but I don't know how to get there.

Any help would be great

Thanks

$\displaystyle \int x^{\frac{3}{2}}e^{-x}dx$ is not elementary, ie, the antiderivative involves functions which are not elementary

• Oct 7th 2009, 02:30 PM
Chloe18
I have the solution but it's not detailed enough for me to understand it

I was able to solve the problem until the step i wrote then got stuck.

It has something to do with kernel of gamma (5/2,1) but i don't know what that means
• Oct 7th 2009, 03:24 PM
artvandalay11
well if you are willing to use the gamma function, evaluate it at z=5/2

Gamma function - Wikipedia, the free encyclopedia
• Oct 7th 2009, 04:05 PM
Chloe18
[quote=artvandalay11;379292]well if you are willing to use the gamma function, evaluate it at z=5/2

why is it 5/2 and not 3/2?
• Oct 7th 2009, 04:08 PM
artvandalay11
[quote=Chloe18;379312]
Quote:

Originally Posted by artvandalay11
well if you are willing to use the gamma function, evaluate it at z=5/2

why is it 5/2 and not 3/2?

You want to evaluate this:
$\displaystyle \int_0^\infty x^{\frac{3}{2}}e^{-x}dx$

And the gamma function is this

$\displaystyle \Gamma (z)=\int_0^\infty t^{z-1}e^{-t}dt$

So $\displaystyle z-1=\frac{3}{2}$

Now I am not aware of a way to evaluate this for non-integer values, not that I'm an expert with the Gamma function, but it seems to me we're back to using tables and calculators
• Oct 7th 2009, 04:34 PM
Chloe18
Oh i get it, thanks

I looked up the value online :)