# Calculate $\int_0^\infty \exp(-\frac{1}{2}x^2)x^k \, \mathrm{d}x$

$$k \in \mathbb N$$. I know, that this is an odd function, if $$k$$ is an even number. So I have to calculate the integral from $$-\infty$$ to $$+\infty$$ and that would be $$0$$ for $$k$$ as an even number. But how do I calculate the integral for an odd number of $$k$$. Wolfram Alpha says something like:

$$\int_0^\infty \exp\left(-\frac{1}{2}x^2\right)x^k \, \mathrm{d}x = 2^{\frac{k-1}{2}} \Gamma\left(\frac{k+1}{2}\right)$$

But how do I get there? Or is there a easier method?

• Sub $x=\sqrt{2 u}$ and use the definition of the Gamma function. – Ron Gordon Nov 11 '15 at 21:53
• Compute $\int e^{-ax^2/2}=\sqrt{\pi/a}$ by squaring and converting to polar coordinates. Then take the derivative of the answer with respect to $a$ $k/2$ times and set $a=1$. – ziggurism Nov 11 '15 at 22:06

Hint: Let $t=\dfrac{x^2}2,~$ and then employ the well-known integral expression for the $\Gamma$ function.