2
$\begingroup$

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!

$\endgroup$
  • $\begingroup$ $s/ax=p, t/bx=q$ seems to be a good start $\endgroup$ – tired Jan 30 '17 at 13:25
  • $\begingroup$ yep afterwards you can integrate one Gaussian followed by two elementary integrations..QED $\endgroup$ – tired Jan 30 '17 at 13:35
  • $\begingroup$ As a general rule of thumb, functions that are defined by an integral can usually be integrated by parts. This is how the integral of $\ln{(x)}$ and $\arctan{(x)}$ and other such functions are derived. $\endgroup$ – David H Jan 30 '17 at 14:10
  • $\begingroup$ Thank you guys for all your answers and suggestions, special thanks to Jack. $\endgroup$ – Mim Jan 30 '17 at 18:09
  • $\begingroup$ You are welcome. $\endgroup$ – Did Apr 9 '17 at 16:18
3
$\begingroup$

At first, we may remove the useless extra parameter, then study $$ I(a) = \int_{0}^{+\infty}\text{erfc}(ax)\,\text{erfc}(x)\,dx $$ through differentiation under the integral sign. Assuming $a>0$ we have $$ I'(a) = \frac{1}{\sqrt{\pi}}\int_{0}^{+\infty}-2x e^{-a^2 x^2}\text{erfc}(x)\,dx \stackrel{pp}{=}\frac{1}{a^2\sqrt{\pi}}-\frac{2}{\pi a^2}\int_{0}^{+\infty}e^{-(a^2+1)x^2}\,dx$$ from which: $$ I'(a) = \frac{1}{\sqrt{\pi}}\left(\frac{1}{a^2}-\frac{1}{a^2\sqrt{1+a^2}}\right) $$ and: $$ I(a) = \frac{1}{\sqrt{\pi}}\left(\frac{\sqrt{1+a^2}-1}{a}\right). $$ Through a substitution, if $a,b>0$ we get: $$ \int_{0}^{+\infty}\text{erfc}(ax)\,\text{erfc}(bx)\,dx = \color{red}{\frac{a+b-\sqrt{a^2+b^2}}{ab\sqrt{\pi}}} $$ as wanted.

$\endgroup$
  • $\begingroup$ And this should solve the issue! Great answer! $\endgroup$ – Mim Jan 30 '17 at 17:52
  • $\begingroup$ @Mim: thank you, you're welcome. $\endgroup$ – Jack D'Aurizio Jan 30 '17 at 18:12
5
$\begingroup$

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 $$I=\frac1{\sqrt{\pi}ab}\left(a-\frac{a^2}{\sqrt{a^2+b^2}}+b-\frac{b^2}{\sqrt{a^2+b^2}}\right)=\frac1{\sqrt{\pi}ab}\left(a+b-\sqrt{a^2+b^2}\right)$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.