If $\displaystyle\int\limits_0^{\infty}e^{-x^2}dx=\sqrt{\pi}/2$, find
$\displaystyle\int\limits_0^{\infty}xe^{-x^2}dx$
Applying by parts, $\displaystyle\int\limits_0^{\infty}xe^{-x^2}dx=\left[x\int e^{-x^2}dx-\int \sqrt{\pi}/2 dx\right]_0^{\infty}=\left[x\int e^{-x^2}dx- \sqrt{\pi}/2 x\right]_0^{\infty}$
Integral calculator says $\displaystyle\int e^{-x^2}dx$ can be simplified to $\dfrac{\sqrt{{\pi}}\operatorname{erf}\left(x\right)}{2}+C$.
${\displaystyle \operatorname {erf} z={\frac {2}{\sqrt {\pi }}}\int _{0}^{z}e^{-t^{2}}\, dt.},$ where $z$ is a complex number. When we simplify $\displaystyle\int _{0}^{z}e^{-t^2}$ we get the same $\operatorname{erf}(x)$ function again, and we're stuck.
However complex functions are not in our syllabus, and I ideally want to solve to this question without using this concept. Someone please help 🙂
Best Answer
Note
\begin{align} \int_0^\infty re^{-r^2}dr=&\frac2\pi \int_0^{\pi/2}\int_0^\infty e^{-r^2}rdrd\theta\\ =&\frac2\pi \int_0^{\infty}\int_0^{\infty} e^{-(x^2+y^2)}dxdy\\ =&\frac2\pi\left(\int_0^{\infty}e^{-x^2}dx\right)^2 = \frac2\pi\cdot \left(\frac{\sqrt{\pi}}2\right)^2=\frac12 \end{align}