1. ## 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

2. 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

3. 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

4. well if you are willing to use the gamma function, evaluate it at z=5/2

Gamma function - Wikipedia, the free encyclopedia

5. [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?

6. [quote=Chloe18;379312]
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

7. Oh i get it, thanks

I looked up the value online