Integral of product of two error complementary functions (erfc) Could you please help me  to show that the integral
$$
\int_0^{\infty} \mathrm{erfc}(ax) \, \mathrm{erfc}(bx)\, \mathrm{d}x
$$
is equal to 
$$
\frac{1}{ab\sqrt{\pi}} (a+b-\sqrt{a^2+b^2}),
$$
where 
$$
\mathrm{erfc}(y)=\frac{2}{\sqrt{\pi}} \int_y^{\infty} \exp(-t^2)\, \mathrm{d} t.
$$
I have tried to expand the integral as
$$
\frac{4}{\pi }\int_0^{\infty} \int_{ax}^{\infty} \int_{bx}^{\infty} \exp(-t^2 -s^2) \, \mathrm{d}s \, \mathrm{d}t \, \mathrm{d} x
$$
but I could not come up with the right change of variables. Any ideas on how to proceed? 
Thank you in advance!


Use the probabilistic way, which shows this is a problem of geometry in the plane...

That is, first note that, for every $x$, $$\mathrm{erfc}(x)=2P(X\geqslant \sqrt2 x)$$ where $X$ is standard normal hence, considering $(X,Y)$ i.i.d. standard normal and assuming that $a$ and $b$ are positive, one sees that the integral to be computed is $$I=4\int_0^\infty P(A_x)dx$$ where $$A_x:=[X\geqslant ax\sqrt2,Y\geqslant bx\sqrt2]$$ Now, $$(X,Y)=(R\cos\Theta,R\sin\Theta)$$ where $(R,\Theta)$ is independent, $\Theta$ is uniform on $(0,2\pi)$ and $R\geqslant0$ is such that $$P(R\geqslant r)=e^{-r^2/2}$$ for every $r\geqslant0$, hence $A_x\subset A_0=[\Theta\in(0,\pi/2)]$ and, on the event $A_0$, $$P(A_x\mid\Theta)=P(R\cos\Theta\geqslant ax\sqrt2,R\sin\Theta\geqslant bx\sqrt2\mid\Theta)=e^{-x^2u(\Theta)^2}$$ with $$u(\theta):=\max\left(\frac{a}{\cos\theta},\frac{b}{\sin\theta}\right)$$ Thus, still on $A_0$, $$\int_0^\infty P(A_x\mid\Theta)dx=\int_0^\infty e^{-x^2u^2(\Theta)}dx=\frac{\sqrt{\pi}}{2u(\Theta)}$$ This proves that $$\sqrt\pi I=4\sqrt\pi E\left(\int_0^\infty P(A_x\mid\Theta)dx\right)=2\pi E\left(\frac1{u(\Theta)}\mathbf 1_{A_0}\right)$$ that is, $$\sqrt\pi I=2\pi E\left(\min\left(\frac{\cos\Theta}a,\frac{\sin\Theta}b\right)\mathbf 1_{A_0}\right)$$ which is, using the distribution of $\Theta$, $$\sqrt\pi I=\int_0^{\pi/2}\min\left(\frac{\cos\theta}a,\frac{\sin\theta}b\right)d\theta=\frac1b\int_0^{\vartheta(a,b)}\sin\theta d\theta+\frac1a\int_{\vartheta(a,b)}^{\pi/2}\cos\theta d\theta$$ where the angle $\vartheta(a,b)$ in $(0,\pi/2)$ is uniquely defined by the condition that $$\tan\vartheta(a,b)=b/a$$ hence $$\sqrt\pi I=\frac{1-\cos\vartheta(a,b)}b+\frac{1-\sin\vartheta(a,b)}a$$ Finally, $$\cos\vartheta(a,b)=\frac{a}{\sqrt{a^2+b^2}}\qquad\sin\vartheta(a,b)=\frac{b}{\sqrt{a^2+b^2}}$$ hence $$I=\frac1{\sqrt{\pi}ab}\left(a-\frac{a^2}{\sqrt{a^2+b^2}}+b-\frac{b^2}{\sqrt{a^2+b^2}}\right)=\frac1{\sqrt{\pi}ab}\left(a+b-\sqrt{a^2+b^2}\right)$$
