How to solve these calculus problems using substitution and the gamma function

[Askhwhelp]

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)?

Please show me the process to get to the answer please
How to use definition of gramma function to solve these?

 

[Dick]

I forgot to do these...long time ago

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

Please show me the process to get to the answer please

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.

 

[Dick]

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

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?

 

[Askhwhelp]

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?

How to use definition of gramma function to solve these?

 

[Dick]

How to use definition of gramma function to solve these?

'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'.

 

[Askhwhelp]

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?

 

[Dick]

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?

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

 

[Askhwhelp]

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

 

[Infrared]

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

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

 

[Askhwhelp]

Does anyone find a earlier way to do the question?

 

[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
$\frac{1}{\sqrt{2}}\int_0^\infty e^{-u}u^{-1/2}du=\frac{\Gamma(1/2)}{\sqrt{2}}$

 

[Dick]

Does anyone find a earlier way to do the question?

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.