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!

A multiple integral approach. Since
$$
\int_{0}^{\infty}x^2 e^{-\alpha x^2}\mathrm{d}x \overset{\alpha x^2 \to x}{=} \frac{1}{2\alpha^{\frac{3}{2}} }\int_{0}^{\infty} x^{\frac{3}{2}-1}e^{-x} \mathrm{d}x = \frac{1}{2\alpha^{\frac{3}{2}} } \Gamma\left(\frac{3}{2}\right)  = \frac{\sqrt{\pi}}{4\alpha^{\frac{3}{2}} } 
$$
We get
\begin{align}
\int_0^\infty \operatorname{erfc}(ax) \operatorname{erfc}(bx)\, \mathrm{d}x & =\int_0^\infty \left( \frac{2}{\sqrt{\pi}}\int_1^\infty ax e^{-a^2x^2z^2} \, \mathrm{d}z\right)\left( \frac{2}{\sqrt{\pi}}\int_1^\infty bx e^{-b^2x^2y^2}\, \mathrm{d}y\right) \mathrm{d}x\\
& =\frac{4ab}{\pi} \int_1^\infty \int_1^\infty \int_0^\infty x^2 e^{-(a^2z^2 +b^2y^2)x^2} \, \mathrm{d}x\, \mathrm{d}z\, \mathrm{d}y\\
& =\frac{ab}{\sqrt{\pi}} \int_1^\infty \int_1^\infty \frac{1}{\left(a^2z^2 +b^2y^2 \right)^{\frac32}} \, \mathrm{d}z\, \mathrm{d}y\\
& \overset{z\to \tan(\theta)by/a}{=}\frac{ab}{\sqrt{\pi}} \int_1^\infty \frac{1}{ab^2y^2} - \frac{1}{y^2b^2\sqrt{y^2b^2+a^2}} \, \mathrm{d}y\\
& =\frac{ab}{\sqrt{\pi}}\left(\frac{1}{ab^2} - \frac{1}{b^2}\int_1^\infty  \frac{1}{y^2\sqrt{b^2y^2+a^2}} \, \mathrm{d}y\right)\\
& \overset{y\to\tan(\varphi)a/b }{=}\frac{a}{b\sqrt{\pi}}\left(\frac{1}{a} - \frac{\sqrt{a^2+b^2} - b}{a^2}  \right)\\
& = \frac{a+b - \sqrt{a^2+b^2}}{ab\sqrt{\pi}}
\end{align}
