If $\int\limits_0^{\infty}e^{-x^2}dx=\sqrt{\pi}/2$, find $\int\limits_0^{\infty}xe^{-x^2}dx$ 
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 :)
 A: 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}
A: As noticed in the comments, we don't need to know the value for the first integral indeed have that
$$\int\limits_0^{\infty}xe^{-x^2}dx=\left[-\frac12e^{-x^2}\right]_0^\infty=\frac12$$
As an alternative, by polar and cartesian integration for the volume of revolution around $y$ axis of the area between the graph $e^{-x^2}$ and $x$ axis, we obtain
$$\int\limits_0^{2\pi}\int\limits_0^{\infty}re^{-r^2}dr d\theta=\int\limits_{-\infty}^{\infty}\int\limits_{-\infty}^{\infty}e^{-(x^2+y^2)}dx dy$$
$$2\pi\int\limits_0^{\infty}re^{-r^2}dr =\int\limits_{-\infty}^{\infty}e^{-y^2}\left(\int\limits_{-\infty}^{\infty}e^{-x^2}dx \right)dy=\sqrt \pi\int\limits_{-\infty}^{\infty}e^{-y^2}dy= \pi$$
that is
$$\int\limits_0^{\infty}re^{-r^2}dr = \frac 12$$
A: I have a much easier solution .
Let $f(x) = e^{-x^2} $ , therefore , $$f'(x) = -2 x e^{-x^2}$$ and hence , $$x e^{-x^2} = \frac{f'(x)}{-2} ....(1)$$
Integrating (1) within limits 0 to $\infty$
$$I=\int_{0}^{\infty} { x e^{-x^2} dx} = {\frac{-1}{2} } \int_{0}^{\infty} f'(x) dx = {\frac{-1}{2} } | f(x) |_{0}^{\infty}  =  \frac{-1}{2} | e^{-x^2} |_{0}^{\infty}$$
And hence , $$I = \int_{0}^{\infty} { x e^{-x^2}}=  (\frac{-1}{2}) (-1) = {1 \over 2}$$
