# Calculus: finding integral

1. Nov 28, 2013

I forgot to do these....long time ago

1) What is integral from 0 to infinity of (y^b * e^(-y/2) dy)?

2) what is the integral from 0 to infinity of (e^(-y^2/2) dy)?

How to use definition of gramma function to solve these?

Last edited: Nov 28, 2013
2. Nov 28, 2013

### Dick

It's a gamma function. If it were long ago and b were a small integer then you would work it out using substitution and integration by parts. I think you are supposed to use the definition of the gamma function.

3. Nov 28, 2013

### Dick

And that's a gaussian integral. It's not considered an elementary integral either. What you learned a long time ago won't help. Isn't your course providing you with any clues?

4. Nov 28, 2013

How to use definition of gramma function to solve these?

5. Nov 28, 2013

### Dick

'gamma' not 'gramma'. And what's the definition of the gamma function? Showing the definition might help you to get started. It won't help for the second one, that's different. Look up 'gaussian integrals'.

6. Nov 29, 2013

For question 1, I use u substitution, u = y/2 so du = 1/2 dy...so integral from 0 to infinity of ((2u)^b*e^-u du) <=> 2^(b+1) * integral from 0 to infinity of (u^b*e^-u du) <=> 2^(b+1) * integral from 0 to infinity (u^((b+1)-1)*e^-u du) <=> 2^(b+1) * gamma symbol which looks like F(b+1)=2^(b+1)* b!
However, the answer my professor gave is 1. I am wondering how could that be?

Last edited: Nov 29, 2013
7. Nov 29, 2013

### Dick

Don't know. That the integral is $2^{b+1} \Gamma(b+1)$ is right. I don't think that's 1.

8. Nov 29, 2013

For question 2, the answer I found is 1/(2*(sqrt(pi))*Γ(1/2).How to find Γ(1/2)?

9. Nov 29, 2013

### Infrared

Are you aware of the reflection identity $\Gamma(z)\Gamma(1-z)=\frac{\pi}{sin(\pi z)}$?

10. Nov 29, 2013

Does anyone find a earlier way to do the question?

11. Nov 29, 2013

### Infrared

I don't see an easier way to do the second question than letting $u=\frac{y^2}{2}$ and solving for $dy$ (which is what I imagine you did) unless you are already familiar with Gaussian integrals.

Edit: I'm not getting what you got. With the substitutions I mentioned, the integral becomes
$\frac{1}{\sqrt{2}}\int_0^\infty e^{-u}u^{-1/2}du=\frac{\Gamma(1/2)}{\sqrt{2}}$

Last edited: Nov 29, 2013
12. Nov 29, 2013

### Dick

As HS-Scientist just said, you don't have to go through the gamma function if you know a simple gaussian integral and use a substitution. And the gaussian you can get in a pretty simple way if you look it up.