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There is this integral that I used a lot in my research: $$\int_{-\infty}^{\infty}\exp\left(-b^{2}\left(x-c\right)^{2}\right)\mathrm{erf}\left(a\left(x-d\right)\right)\,\mathrm{d}x = \frac{\sqrt{\pi}}{b}\mathrm{erf}\left(\frac{ab\left(c-d\right)}{\sqrt{a^{2}+b^{2}}}\right),$$ for $a,b>0$.

It can be evaluated using the "differentiation under integral sign" method as follows: $$ {\small \begin{aligned} I\left(d\right)&=\int_{-\infty}^{\infty}\exp\left(-b^{2}\left(x-c\right)^{2}\right)\mathrm{erf}\left(a\left(x-d\right)\right)\,\mathrm{d}x\\ \frac{\mathrm{d}}{\mathrm{d}d}I\left(d\right)&=\frac{\mathrm{d}}{\mathrm{d}d}\int_{-\infty}^{\infty}\exp\left(-b^{2}\left(x-c\right)^{2}\right)\mathrm{erf}\left(a\left(x-d\right)\right)\,\mathrm{d}x\\ &= \int_{-\infty}^{\infty}\frac{\mathrm{\partial}}{\partial d}\exp\left(-b^{2}\left(x-c\right)^{2}\right)\mathrm{erf}\left(a\left(x-d\right)\right)\,\mathrm{d}x\\ &\overset{\mathtt{M}}{=} -\frac{2a}{\sqrt{\pi}}\int_{-\infty}^{\infty}\exp\left(-b^{2}\left(x-c\right)^{2}-a^{2}\left(x-d\right)^{2}\right)\,\mathrm{d}x\\ &=-\frac{2a}{\sqrt{\pi}}\int_{-\infty}^{\infty}\exp\left(-\left(a^{2}+b^{2}\right)\left(x-\frac{b^{2}c+a^{2}d}{a^{2}+b^{2}}\right)^{2}+\frac{\left(b^{2}c+a^{2}d\right)^{2}}{a^{2}+b^{2}}-\left(b^{2}c^{2}+a^{2}d^{2}\right)\right)\,\mathrm{d}x\\ & =-\frac{2a}{\sqrt{\pi}}\exp\!\left(\frac{\left(b^{2}c+a^{2}d\right)^{2}}{a^{2}+b^{2}}-\left(b^{2}c^{2}+a^{2}d^{2}\right)\right)\int_{-\infty}^{\infty}\exp\left(-\left(a^{2}+b^{2}\right)\left(x-\frac{b^{2}c+a^{2}d}{a^{2}+b^{2}}\right)^{2}\right)\,\mathrm{d}x\\ &\overset{\mathtt{*}}{=}-\frac{2a}{\sqrt{\pi}}\exp\left(-\frac{a^{2}b^{2}}{a^{2}+b^{2}}\left(c-d\right)^{2}\right)\int_{-\infty}^{\infty}\exp\left(-\left(a^{2}+b^{2}\right)y^{2}\right)\,\mathrm{d}y\\ & \overset{\mathtt{M}}{=}-\frac{2a}{\sqrt{a^{2}+b^{2}}}\exp\left(-\frac{a^{2}b^{2}\left(c-d\right)^{2}}{a^{2}+b^{2}}\right). \end{aligned} } $$ $\overset{\mathtt{*}}{=}$ substitution $y=x-\frac{b^{2}c+a^{2}d}{a^{2}+b^{2}}$

$\overset{\mathtt{M}}{=}$ calculated with Mathematica

Thus

$$ \begin{aligned} I\left(d\right)&=-\frac{2a}{\sqrt{a^{2}+b^{2}}}\int\exp\left(-\frac{a^{2}b^{2}\left(c-d\right)^{2}}{a^{2}+b^{2}}\right)\,\mathrm{d}d\\ & \overset{\mathtt{M}}{=}\frac{\sqrt{\pi}}{b}\mathrm{erf}\left(\frac{ab\left(c-d\right)}{\sqrt{a^{2}+b^{2}}}\right)+C. \end{aligned} $$

And it can be easily shown that $C=0$.

Now, I would like to evaluate a similar integral with different limits:

$$ \int_{0}^{\infty}\exp\left(-b^{2}\left(x-c\right)^{2}\right)\mathrm{erf}\left(a\left(x-d\right)\right)\,\mathrm{d}x. $$

It seems to me that the first obvious difference is doing $\int_{0}^{\infty}\exp\left(-\left(a^{2}+b^{2}\right)y^{2}\right)\,\mathrm{d}y = \frac{\sqrt\pi}{2\sqrt{a^{2}+b^{2}}}$ instead of $ \int_{-\infty}^{\infty}\exp\left(-\left(a^{2}+b^{2}\right)y^{2}\right)\,\mathrm{d}y =\frac{\sqrt\pi}{\sqrt{a^{2}+b^{2}}}$ in the line starting with $\overset{\mathtt{*}}{=}$.

Also, assuming $d\rightarrow+\infty$, $C = -\frac{\sqrt{\pi }}{2 b}\mathrm{erf}\left(bc\right)$.

In conclusion, the integral in question should evaluate to $$ \int_{0}^{\infty}\exp\left(-b^{2}\left(x-c\right)^{2}\right)\mathrm{erf}\left(a\left(x-d\right)\right)\,\mathrm{d}x = \frac{\sqrt{\pi}}{2b}\mathrm{erf}\left(\frac{ab\left(c-d\right)}{\sqrt{a^{2}+b^{2}}}\right) -\frac{\sqrt{\pi }}{2 b}\mathrm{erf}\left(bc\right) $$

but it is not correct.

So, does the differentiation under integral sign method fail for some reason in this case, or am I making some silly mistake?

Thanks.

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  • $\begingroup$ The substitution $y=x-something$ changes the bounds of integration, at the end of your calculation. $\endgroup$ – Omran Kouba Apr 21 '14 at 12:38
  • $\begingroup$ @OmranKouba Right, so it was a silly mistake after all. But now it turns out I cannot evaluate the integral. I have been struggling with similar integrals for quite some time now and I'm out of ideas. Is there some kind of explaination why the integral with infinite limits "works", but when I change the lower limit to zero, it seems undoable. Maybe it is not a valid question but I am really curious since I am not a trained mathematician. $\endgroup$ – petru Apr 21 '14 at 13:04

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