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Can anybody tell me how to prove the following integral identities (1) and (2) involving error function erf?

(1) $\int_{0}^{t}\int_{0}^{\infty}\frac{\exp\left[-\frac{\left(x-\xi\right)^{2}}{4D\left(t-\tau\right)}\right]-\exp\left[-\frac{\left(x+\xi\right)^{2}}{4D\left(t-\tau\right)}\right]}{\sqrt{4\pi D\left(t-\tau\right)}}erf\left(\frac{\xi}{\sqrt{4D\tau}}\right)\, d\xi\, d\tau=t\, erf\left(\frac{x}{\sqrt{4Dt}}\right)$.

It is assumed that $D>0$ and $x\geq0$, and the internal integration is along $\xi$. MATHEMATICA function Integrate[] fails to calculate the above integral symbolically (there is a question why?), but numerical evaluation confirms the identity.

I would also be happy to be able to calculate analytically (or prove that this is impossible) the integral $\int_{0}^{t}\int_{0}^{\infty}\frac{\exp\left[-\frac{\left(x-\xi\right)^{2}}{4D\left(t-\tau\right)}\right]-\exp\left[-\frac{\left(x+\xi\right)^{2}}{4D\left(t-\tau\right)}\right]}{\sqrt{4\pi D\left(t-\tau\right)}}\left[erf\left(\frac{\xi}{\sqrt{4D\tau}}\right)\right]^{2}\, d\xi\, d\tau$.

(2) $\int_{0}^{\infty}\exp\left(-\xi^{2}a^{2}\right)erf\left(\xi b\right)\, d\xi=\frac{\arctan\left(b/a\right)}{a\sqrt{\pi}}$,

where $a>0$ and $b>0$. This result is returned by MATHEMATICA, but I would be glad to know how it is obtained.

Leslaw

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1 Answer 1

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For (2), you can use the Maclaurin series for $\text{erf}(\xi b)$ and the fact that $$\int_0^\infty \exp(-\xi^2 a^2) \xi^{2k+1}\; d\xi = \frac{k!}{2 a^{2k+2}}$$

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