Expected value of $X^n$ for normal distribution Let there be a random variable $X$ and $N(0, \lambda ^2)$ given.
I am to find $E[X^n]$ knowing that $\Gamma \big(\frac{1}{2} \big) = \frac{\pi}{2}$. 

I started with the definition:
$$E[X^n] = \frac{1}{\sqrt{2 \pi}\lambda}\int \limits_{- \infty}^{\infty}x^ne^{\frac{-x^2}{2 \lambda^2}} \mbox{d}x$$
I figured out that:
$$E[X^n] = \begin{cases} 0 &\text{for } n = 2k + 1; k \in \mathbb{N} \cup\{0\}\\\text{something}&\text{for } n = 2k; k \in \mathbb{N} \end{cases}$$

So now:
$$ \frac{1}{\sqrt{2 \pi}\lambda}\int \limits_{- \infty}^{\infty}x^ne^{\frac{-x^2}{2 \lambda^2}} \mbox{d}x =  \frac{2}{\sqrt{2 \pi}\lambda}\int \limits_{0}^{\infty}x^ne^{\frac{-x^2}{2 \lambda^2}} \mbox{d}x$$
I substitute:
$$t = \frac{x}{\lambda}$$
$$\mbox{d}x = \lambda \mbox{d}t$$ 
Thus:
$$\frac{2}{\sqrt{2 \pi}\lambda}\int \limits_{0}^{\infty}x^ne^{\frac{-x^2}{2 \lambda^2}} \mbox{d}x = \frac{2 \lambda^{n}}{\sqrt{2 \pi}}\int \limits_{0}^{\infty}t^ne^{\frac{-t^2}{2}} \mbox{d}t.$$
By integrating the formula above by parts I do get: 
$$\frac{2 \lambda^{n}}{\sqrt{2 \pi}}(n-1) \int \limits_{0}^{\infty}t^{n-2}e^{\frac{-t^2}{2}} \mbox{d}t$$
I don't really know where to go from here. And unfortunately I don't see how can I use the Gamma Function, which is given by this formula:
$$\Gamma (z) = \int \limits_{0}^{\infty} x^{z-1}e^{-x} \mbox{d}x,$$
isn't it?
 A: First of all, a correction: the value of $\Gamma(\frac12)$ is $\sqrt{\pi}$, not $\frac\pi2$.
You've done most of the work. If you inspect the identity that you obtained through integration by parts:
$$
E(X^n)=\frac{2 \lambda^{n}}{\sqrt{2 \pi}}\int \limits_{0}^{\infty}t^ne^{\frac{-t^2}{2}} \mbox{d}t=\frac{2 \lambda^{n}}{\sqrt{2 \pi}}(n-1) \int \limits_{0}^{\infty}t^{n-2}e^{\frac{-t^2}{2}} \mbox{d}t\tag1
$$
you've just proved the following:
$$
E(X^n)=(n-1)\lambda^2E(X^{n-2}).\tag2
$$
Now (2) is a recursive formula for generating $E(X^n)$ when $n$ is even. To apply this formula you need to get going with the value of $E(X^n)$ for some starting (base) value of $n$, such as $n=0$. But when $n=0$, you're computing the expectation of the constant $1$, which is $1$. Applying (2) repeatedly to this base case, we obtain:
$$\begin{align} E(X^0)&=1, \\
E(X^2)&=(2-1)\lambda^2E(X^0)=\lambda^2, \\ 
E(X^4)&=(4-1)\lambda^2E(X^2)=3\lambda^4, \\ 
E(X^6)&=(6-1)\lambda^2E(X^4)=15\lambda^6, \end{align}$$
and so on. Can you see the general formula?
It wasn't necessary to use the Gamma function, though you can use the substitution $x=t^2/2$ in the LHS of (1) to obtain a closed-form expression for $E(X^n)$ in terms of Gamma:
$$
E(X^n)=\frac{2^{n/2}\lambda^n}{\sqrt\pi}\Gamma\left(\frac{n+1}2\right).\tag3
$$
You can plug $n=0$ into (3) to find
$$
E(X^0)=\frac1{\sqrt\pi}\Gamma(\textstyle\frac12)=1,
$$
but you don't need the Gamma function to know that $E(X^0)$ is $1$.
