How to evaluate $\int_{0}^{\infty} x^4 e^{-0.5x^2} dx$

Question

How to evaluate $\int_{0}^{\infty} x^4 e^{-0.5x^2} dx$?

I am trying to evaluate this integral, but to no avail. I am completely stuck and have tried all the classic methods (u-substitution, integration by parts, etc). Could someone point me in the right direction? It has been ages since I took calculus.

Answer 1

Hint Using the symmetry of the integrand and applying $u$-substitution to the Gaussian integral gives: $$\int_0^\infty e^{-a^2 x^2} \,dx = \frac{\sqrt\pi}{2 a} .$$

Differentiating with respect to $a$ gives $$-2 a\int_0^\infty x^2 e^{-a^2 x^2} \,dx = -\frac{\sqrt\pi}{2 a^2} ,$$ and then rearranging gives $$\int_0^\infty x^2 e^{-a^2 x^2} \,dx = \frac{\sqrt\pi}{4 a^3} .$$

Now, differentiate again.

We get $$-2 a\int_0^\infty x^4 e^{-a^2 x^2} \,dx = -\frac{3\sqrt\pi}{4 a^4},$$ so $$\boxed{\int_0^\infty x^4 e^{-a^2 x^2} \,dx = \frac{3\sqrt\pi}{8 a^5}} .$$ Taking $a = \frac{1}{\sqrt{2}}$ gives $$\boxed{\int_0^\infty x^4 e^{-\frac{1}{2} x^2} \,dx = 3 \sqrt\frac{\pi}{2}} .$$

More generally we have

$$\int_0^\infty x^k e^{-a^2 x^2} \,dx = \frac{1}{2 a^{k + 1}} \int_0^\infty t^\frac{k - 1}{2} e^{-t} = \frac{1}{2 a^{k + 1}} \Gamma\left(\frac{k + 1}{2}\right) ,$$ where $\Gamma$ denotes the gamma function