How to evaluate $\int_{0}^{\infty} x^4 e^{-x^2/{\beta}^2}dx$? Can someone show me how to evaluate
$\int_{0}^{\infty} x^4 e^{-x^2/{\beta}^2}dx$?
I am working on verifying that a function is a pdf, and finding its expectation and variance. I want to be able to do it step-by-step, without just getting the answer from a CAS. Thanks in advance!
 A: For general use, and when not in need to show step-by-step calculations, note that your integral is a Mellin transform of $e^{-x^2/{\beta}^2}$.
The following general relation holds:
$$\int_0^{\infty} x^{s-1}\exp\left\{-ax^h\right\}dx = h^{-1}a^{- s/h} \Gamma\left( s/h\right)\;, \qquad h>0, \;\text {Re}\,a>0,\;\text {Re}\,s>0 $$
...where $\Gamma()$ is the Gamma function.
This leads of course to the answer given by @Robert Smith, using the specific numerical values.
Finally , you cannot "verify" that it is a pdf - what you can do it to determine for what value of $\beta$ it becomes a proper pdf over the specific domain, i.e. that it integrates to unity.
A: If you know that $$\int_{0}^{\infty} e^{-\alpha x^{2}} dx =  \frac{1}{2}\sqrt{\frac{\pi}{\alpha}}$$ then by differentiating twice under the integral sign, you have:
$$\frac{\partial^{2}}{\partial \alpha^{2}} \int_{0}^{\infty} e^{-\alpha x^{2}} dx =  \int_{0}^{\infty} x^{4} e^{-\alpha x^{2}} dx$$
Therefore, applying the differentiation to the result you already know:
$$\frac{\partial^{2}}{\partial \alpha^{2}} \int_{0}^{\infty} e^{-\alpha x^{2}} dx = \frac{\partial^{2}}{\partial \alpha^{2}} \frac{1}{2}\sqrt{\frac{\pi}{\alpha}} = \frac{3\sqrt{\pi}}{8\alpha^{5/2}}$$
In your problem, simply substitute $\alpha = 1/\beta^{2}$ and you have the result that you can verify in W|A:
$$\int_{0}^{\infty} x^{4} e^{-\alpha x^{2}} dx = \frac{3\sqrt{\pi}}{8\left(\frac{1}{\beta^{2}}\right)^{5/2}}$$
Given your recent update, let me add the following:
You should know that the expectation and variance are given by 
$$E[X] = \int x f(x)dx$$
and $$Var(X) = \int (x-\mu)^2f(x)dx$$
I'm sure you can integrate by substitution the first one. The variance is a bit easier once you know how to apply the Gaussian integral above, since it involves a term $\int x^{2} f(x) dx$ and a constant $\mu$. In that case, of course, you only need to differentiate once.
