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

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- Oct 7th 2009, 02:20 PMChloe18Help 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 PMartvandalay11
- Oct 7th 2009, 02:30 PMChloe18
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 PMartvandalay11
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 PMChloe18
[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 PMartvandalay11
[quote=Chloe18;379312]

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 PMChloe18
Oh i get it, thanks

I looked up the value online :)