Any simple way for proving $\int_{0}^{\infty} \mathrm{erf(x)erfc{(x)}}\, dx = \frac{\sqrt 2-1}{\sqrt\pi}$? How to prove
 $$\int_{0}^{\infty} \mathrm{erf(x)erfc{(x)}}\, dx = \frac{\sqrt 2-1}{\sqrt\pi}$$ with  $\mathrm{erfc(x)} $ is the complementary error function, I have used integration by part but i don't succed 
 A: The given integral equals
$$ \frac{4}{\pi}\int_{0}^{+\infty}\int_{0}^{x}e^{-a^2}\,da \int_{x}^{+\infty}e^{-b^2}\,db\,dx =\frac{4}{\pi}\iiint_{0\leq a\leq x\leq b} e^{-(a^2+b^2)}\,da\,db\,dx$$
or
$$\frac{4}{\pi}\iint_{0\leq a\leq b}(b-a)e^{-(a^2+b^2)}\,da\,db = \frac{4}{\pi}\int_{0}^{+\infty}\int_{0}^{\pi/4}(\cos\theta-\sin\theta)\rho^2 e^{-\rho^2}\,d\theta \,d\rho$$
or
$$ \frac{4}{\pi}(\sqrt{2}-1)\int_{0}^{+\infty}\rho^2 e^{-\rho^2}\,d\rho = \frac{4}{\pi}(\sqrt{2}-1)\frac{\sqrt{\pi}}{4}=\color{red}{\frac{\sqrt{2}-1}{\sqrt{\pi}}}.$$
A: Recalling that $\text{erf} (x) = 1 - \text{erfc} (x)$, the integral can be rewritten as
$$\int_0^\infty \text{erf}(x) \text{erfc}(x) \, dx = \int_0^\infty \text{erfc}(x) \, dx - \int_0^\infty \text{erfc}^2 (x) \, dx.$$
As
$$\frac{d}{dx} \left (\text{erfc}(x) \right ) = -\frac{2}{\sqrt{\pi}} e^{-x^2},$$
integrating by parts gives
\begin{align*}
\int_0^\infty \text{erf}(x) \text{erfc}(x) \, dx &= - \frac{2}{\sqrt{\pi}} \int_0^\infty x e^{-x^2} \, dx + \frac{4}{\sqrt{\pi}} \int_0^\infty x e^{-x^2} \text{erfc}(x) \, dx.
\end{align*}
And by parts again
\begin{align*}
\int_0^\infty \text{erf}(x) \text{erfc}(x) \, dx &= \frac{1}{\sqrt{\pi}} - \frac{4}{\sqrt{\pi}} \left (\frac{1}{2} - \frac{1}{\sqrt{\pi}} \int_0^\infty e^{-2x^2} \, dx \right )\\
&= -\frac{1}{\sqrt{\pi}} + \frac{4}{\pi} \int_0^\infty e^{-2x^2} \, dx.
\end{align*}
In the last integral, enforcing a substitution of $x \mapsto x/\sqrt{2}$ leads to
\begin{align*}
\int_0^\infty \text{erf}(x) \text{erfc}(x) \, dx &= -\frac{1}{\sqrt{\pi}} + \frac{4}{\pi \sqrt{2}} \frac{\sqrt{\pi}}{2} \cdot \frac{2}{\sqrt{\pi}} \int_0^\infty e^{-x^2} \, dx\\
&= -\frac{1}{\sqrt{\pi}} + \frac{2}{\sqrt{\pi} \sqrt{2}},
\end{align*}
or
$$\int_0^\infty \text{erf}(x) \text{erfc}(x) \, dx = \frac{\sqrt{2} - 1}{\sqrt{\pi}},$$
as expected.
