3
$\begingroup$

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

| cite | improve this question | | | | |
$\endgroup$
  • 2
    $\begingroup$ Wolfram Alpha gives a primitive. $\endgroup$ – gammatester Feb 13 '18 at 19:20
7
$\begingroup$

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

| cite | improve this answer | | | | |
$\endgroup$
  • $\begingroup$ Thanks for your attention just wrong typo, I meant sqrt(pi) in denominato $\endgroup$ – zeraoulia rafik Feb 13 '18 at 19:22
5
$\begingroup$

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.

| cite | improve this answer | | | | |
$\endgroup$

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.