Integration to Gamma?

Question

So my teacher is integrating to find the expectation of a marginal distribution: $$E(Y) = \int_0^\infty (y) \frac{1+y}{2} e^{-y} dy$$ and goes straight to this: $$\int_0^\infty \frac{y}{2}e^{-y}dy +\int_0^\infty \frac{y^2}{2}e^{-y}dy$$ which he then simplifies to $$\frac{1}{2}\Gamma(2) + \frac{1}{2}\Gamma(3) = \frac{3}{2}$$ I'm not really sure what's happening between the second and third step. I know that $\Gamma(n) = (n-1)!$ so I'm familiar on the last bit of arithmatic, but I'm not sure how he simplifies to the gammas in the last step.

Answer 1

We have

$$\Gamma(x) = \int_{0}^{\infty} t^{x-1} e^{-t} dt$$

In particular

$$n! = \int_0^{\infty} y^n e^{-y} dy$$

which can be justified with integration by parts, without mentioning the Gamma function at all.

Answer 2

$\Gamma (t) = \int_{0}^{\infty} y^{t-1}e^{-y}dy$.

So $(1/2)\Gamma (2) = \int_{0}^{\infty} (1/2)ye^{-y}dy$. And similarly for $(1/2)\Gamma(3)$.