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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

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    $\begingroup$ Wolfram Alpha gives a primitive. $\endgroup$ – gammatester Feb 13 '18 at 19:20
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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 $$ \frac{4}{\pi}(\sqrt{2}-1)\int_{0}^{+\infty}\rho^2 e^{-\rho^2}\,d\rho = \frac{4}{\pi}(\sqrt{2}-1)\frac{\sqrt{\pi}}{4}=\color{red}{\frac{\sqrt{2}-1}{\sqrt{\pi}}}.$$

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

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