Integral $\int_0^{\infty} r^2 \exp (-\frac{r^2}{2}) dr$

Question

Consider $$\int_0^{\infty} r^2 \exp (-\frac{r^2}{2})\,dr.$$ I have tried several times by parts but still do not get it. I know the solution is $\frac{\sqrt{\pi}}{\sqrt{2}}$ but I do not know how to get it.

Any hint?

Answer 1

Integrating by parts is a good idea. A primitive of $re^{-\frac{r^2}2}$ is $-e^{-\frac{r^2}2}$. So,\begin{align}\int_0^\infty r^2e^{-\frac{r^2}2}&=\left[-re^{-\frac{r^2}2}\right]_{r=0}^{r\to\infty}+\int_0^\infty e^{-\frac{r^2}2}\,\mathrm dr\\&=\frac{\sqrt\pi}{\sqrt2}.\end{align}

Answer 2

Let $u=r$, $v'=r\exp(-\frac{r^2}{2})$. Then we know that, by integration by parts,

$$\int_0^{\infty} r^2 \exp (-\frac{r^2}{2})\,dr.=\int_0^\infty uv\,dr=[uv]_0^\infty-\int_0^\infty u'v\, dr$$

which is just

$$-\int_0^\infty v\, dr=\int_0^\infty e^{-\frac{r^2}{2}}\, dr=\sqrt{\frac{\pi}{2}}$$

Answer 3

Since $\int_0^\infty\exp(-\alpha r^2)dr=\frac12\pi^{1/2}\alpha^{-1/2}$, applying $-\partial_\alpha$ gives $\int_0^\infty r^2\exp(-\alpha r^2)dr=\frac14\pi^{1/2}\alpha^{-3/2}$. With $\alpha=\frac12$, your integral is $\frac14\pi^{1/2}2^{3/2}=\sqrt{\frac{\pi}{2}}$.

Answer 4

The integral $I=\int_0^{\infty} r^2\exp (-\frac{r^2}{2})\,dr$ can be evaluated as a double integral: $$1\cdot\sqrt{\frac{\pi}{2}}=\int_0^{\infty} x \exp (-\frac{x^2}{2})dx \cdot \int_0^{\infty} \exp (-\frac{y^2}{2})dy \\=\int_0^{\pi/2}\cos(\theta)d\theta\int_0^{\infty} r^2\exp (-\frac{r^2}{2})dr=1\cdot I.$$ P.S. Recall that the non-trivial integral $J=\int_0^{\infty} \exp (-\frac{y^2}{2})dy$ can be evaluated in a similar way: $$J^2=\int_0^{\infty} \exp (-\frac{x^2}{2})dx\cdot \int_0^{\infty} \exp (-\frac{y^2}{2})dy\\= \int_0^{\pi/2}d\theta\int_0^{\infty} r\exp (-\frac{r^2}{2})dr=\pi/2\cdot 1.$$