integration - Any simple way for proving $int_{0}^{infty} mathrm{erf(x)erfc{(x)}}, dx = frac{sqrt 2-1}{sqrtpi}$?

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

Answer

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 $$ \frac{4}{\pi}(\sqrt{2}-1)\int_{0}^{+\infty}\rho^2 e^{-\rho^2}\,d\rho = \frac{4}{\pi}(\sqrt{2}-1)\frac{\sqrt{\pi}}{4}=\color{red}{\frac{\sqrt{2}-1}{\sqrt{\pi}}}.$$