Calculating gaussian integral

Question

I have a basic idea about Gaussian integral i.e $$\int_{-\infty}^{\infty} e^{-x^2}\,dx = \sqrt {\pi}.$$

How can I calculate this integral using this basic idea: $$\int_{-\infty}^{\infty} x^4e^{-4x^2}\,dx$$

Answer 1

We have : \begin{aligned}\int_{-\infty}^{+\infty}{x^{4}\,\mathrm{e}^{-4x^{2}}\,\mathrm{d}x}&=\underbrace{\left[-\frac{\mathrm{e}^{-4x^{2}}}{8}\times x^{3}\right]_{-\infty}^{+\infty}}_{0}+\frac{3}{8}\int_{-\infty}^{+\infty}{x^{2}\,\mathrm{e}^{-4x^{2}}\,\mathrm{d}x}\\ &=\frac{3}{8}\underbrace{\left[-\frac{\mathrm{e}^{-4x^{2}}}{8}\times x\right]_{-\infty}^{+\infty}}_{0}+\frac{3}{64}\int_{-\infty}^{+\infty}{\mathrm{e}^{-4x^{2}}\,\mathrm{d}x}\\ &=\frac{3}{128}\int_{-\infty}^{+\infty}{\mathrm{e}^{-\left(2x\right)^{2}}\times2\,\mathrm{d}x}\\ &=\frac{3}{128}\int_{-\infty}^{+\infty}{\mathrm{e}^{-y^{2}}\,\mathrm{d}y}\\ \int_{-\infty}^{+\infty}{x^{4}\,\mathrm{e}^{-4x^{2}}\,\mathrm{d}x}&=\frac{3\sqrt{\pi}}{128}\end{aligned}

Note : We did a first integration by parts setting $ u':x\mapsto x\,\mathrm{e}^{-4x^{2}} $, and $ v:x\mapsto x^{3} $, then a second integration by parts setting $ f'=u' $, and $ g=\mathrm{Id}_{\mathbb{R}} $.

Answer 2

Hint:

Consider the integral

$$\int_{\mathbb{R}} e^{-\alpha x^2}\,dx.$$

You can calculate this as a function of $\alpha$ based on your known integral (do a simple change of variable to get it into your original form). What happens if you differentiate this twice with respect to $\alpha$? Can you take it from here?

Answer 3

Hint:

Decompose the integrand as

$$x^3\cdot xe^{-x^2}$$ and integrate by parts.

Then repeat.