Question

Prove ∫^∞₀ 𝑒 ^{−𝑥 2} 𝑥 ² 𝑑𝑥 = √𝜋/ 4

Answer 1

 

$\int \mathrm{e}^{-x^{2}}x^{2} \mathrm{~d} x=$ $\int x^{2} \mathrm{e}^{-x^{2}} \mathrm{~d} x$

Using Integration by parts

 $=-\frac{x \mathrm{e}^{-x^{2}}}{2}-\int-\frac{\mathrm{e}^{-x^{2}}}{2} \mathrm{~d} x$ ...(1)

Now solving:

$\int-\frac{\mathrm{e}^{-x^{2}}}{2} \mathrm{~d} x$

Apply linearity:

 $\begin{aligned} &=-\frac{\sqrt{\pi}}{4} \int \frac{2 \mathrm{e}^{-x^{2}}}{\sqrt{\pi}} \mathrm{d} x \\ &=-\frac{\sqrt{\pi} \operatorname{erf}(x)}{4} \end{aligned}$

Put the above value in eqn .(1)

 $\begin{aligned} &-\frac{x \mathrm{e}^{-x^{2}}}{2}-\int-\frac{\mathrm{e}^{-x^{2}}}{2} \mathrm{~d} x \\ &=\frac{\sqrt{\pi} \operatorname{erf}(x)}{4}-\frac{x \mathrm{e}^{-x^{2}}}{2}+C \end{aligned}$

So, $\int x^{2} \mathrm{e}^{-x^{2}} \mathrm{~d} x=\frac{\sqrt{\pi} \operatorname{erf}(x)}{4}-\frac{x \mathrm{e}^{-x^{2}}}{2}+C$

And $\int_{0}^{\infty} x^{2} \mathrm{e}^{-x^{2}} \mathrm{~d} x=[\frac{\sqrt{\pi} \operatorname{erf}(x)}{4}-\frac{x \mathrm{e}^{-x^{2}}}{2}+C]_{0}^{\infty}=\frac{\sqrt{\pi}}{4}$