# Evaluating $\int_0^\infty x^2e^{-\alpha x^2}dx$ and $\int_0^\infty xe^{-\alpha x^2} dx$ knowing $\int_0^\infty e^{-\alpha x^2}dx$ [closed]

As the title, question 5 in this picture

https://i.stack.imgur.com/P52hf.jpg

thanks

If you know that $$\int_0^\infty e^{-\alpha x^2} \: dx=\frac12\sqrt{\frac{\pi}{\alpha}}, \qquad \alpha>0.$$ Then you may deduce, using an integration by parts, that

\begin{align} I_2(\alpha):= \int_0^\infty x^2e^{-\alpha x^2} \: dx&= \left.\left(-\frac{1}{2\alpha}\right)x \times\left( e^{-\alpha x^2}\right)\right|_0^{\infty}+\frac{1}{2\alpha}\int_0^\infty e^{-\alpha x^2} \: dx=\frac{1}{4\alpha}\sqrt{\frac{\pi}{\alpha}}. \end{align}

The other integral $$I_1(\alpha):=\int_0^\infty xe^{-\alpha x^2} \: dx$$ is easily deduced by a change of variable, setting $u=x^2$, $du=2 xdx$, you get

\begin{align} I_1(\alpha)= \int_0^\infty xe^{-\alpha x^2} \: dx&=\frac12\int_0^\infty e^{-\alpha u} \: du=\frac12\left.\left(-\frac{1}{\alpha}\right) \times\left( e^{-\alpha u}\right)\right|_0^{\infty}=\frac{1}{2\alpha}. \end{align}

• Thanks you very much,your calculus level is extremely high,haha Jul 11, 2015 at 14:04
• @RichardTsai You are welcome! Jul 11, 2015 at 14:12

you might generate arbritary even powers of $x$ by just noticing that

$$\int_{0}^{\infty}x^{2n}e^{-\alpha x^2}=(-1)^n\partial^n_{\alpha}\int_{0}^{\infty}e^{-\alpha x^2}=(-1)^n \frac{\sqrt{\pi}}{2}\partial^n_{\alpha}\sqrt{\frac{1}{\alpha}}$$

But it is not possible to generate odd powers out of the knowledge that $\int_{0}^{\infty}e^{-\alpha x^2}=\frac{1}{2}\sqrt{\frac{\pi}{\alpha}}$!

You need another starting point like

$$\int_{0}^{\infty}xe^{-\alpha x^2}=\frac{-1}{2\alpha}\int_{0}^{\infty}\partial_{x}e^{-\alpha x^2}=\frac{-1}{2\alpha}\left[e^{-\alpha x^2}\right]_{0}^{\infty}=\frac{1}{2\alpha}$$

The you might procced as above

$$\int_{0}^{\infty}x^{2n+1}e^{-\alpha x^2}=(-1)^n\partial^n_{\alpha}\int_{0}^{\infty}xe^{-\alpha x^2}=(-1)^n\frac{1}{2}\partial^n_{\alpha}\frac{1}{\alpha}$$

Check that if you let say $u=ax^2$ then $dx=du/(2ax)$ but $x=\sqrt{(u/a)}$ substituting in the integral and simplify it reduces to $(1/(2a^{1.5})) \gamma(3/2)$and further reduces to $(1/4)(1/a^{1.5})\sqrt{\pi}$. Or check integrals involving gamma functions.