Calculus: finding integral I need to compute the following two integrals: $$\int_0^{\infty}y^be^{-y/2} dy$$
$$\int_0^{\infty}\frac{1}{\sqrt{2\pi}}e^{-y^2/2}dy$$  can I do this?  Please show me the process.
How to use definition of gamma function to solve these?
 A: For the first integral, substitute $u=y/2$ with $2du=dy$ to get:
$$ 2^{b+1}\int_0^{\infty}u^be^{-u}du = 2^{b+1}\Gamma(b+1)$$
from the definition of the gamma function.
The second integral is a Gaussian integral, so I will not try to replicate what has been done so well before.
EDIT:
I see you want to use the definition of the gamma function.
For the second integral, let $u=y^2/2$ with $du=ydy$ to get:
$$\frac{1}{2\sqrt{\pi}}\int_0^\infty u^{-1/2}e^{-u}du=\frac{1}{2\sqrt{\pi}}\Gamma(1/2)$$
Since $\Gamma(1/2)=\sqrt{\pi}$, you get $1/2$ as a final answer. Of course finding $\Gamma(1/2)$ is another story.
Finding $\Gamma(1/2)$
The most logical way in your case is to go back, evaluate the second integral as a Gaussian integral, and then compare results. You can find online many good derivations of the Gaussian integral. I found this one well-explained.
But if you assume known the reflection property of the gamma function:
$$\Gamma(1-z) \Gamma(z) = \frac{\pi}{\sin \pi z}$$
you can set $z = 1/2$ to get $\Gamma(1/2)^2=\pi$ or $\Gamma(1/2) = \sqrt{\pi}$.
