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

Question

I have big difficulties solving the following integral: $$ \int_{-\infty}^{\infty}x\exp\left(-b^{2}\left(x-c\right)^{2}\right)\mathrm{erf}^{2}\left(a\left(x-d\right)\right)\,\mathrm{d}x $$

I tried to use integration by parts, and also tried to apply the technique called “differentiation under the integration sign” but with no results.

I’m not very good at calculus so my question is if anyone could give me any hint of how to approach this integral. I would be ultimately thankful.

If it could help at all, I know that $$ \int_{-\infty}^{\infty}x\exp\left(-b^{2}\left(x-c\right)^{2}\right)\mathrm{erf}\left(a\left(x-d\right)\right)\,\mathrm{d}x=\frac{a}{b^{2}\sqrt{a^{2}+b^{2}}}\exp\left(-\frac{a^{2}b^{2}\left(c-d\right)^{2}}{a^{2}+b^{2}}\right)+\frac{\sqrt{\pi}c}{b}\mathrm{erf}\left(\frac{ab\left(c-d\right)}{\sqrt{a^{2}+b^{2}}}\right), $$

for $b>0$.

Answer 1

A suggestion instead of a complete answer: taking $c=0$ for ease of typing and integrating by parts $$\begin{align*} \int_{-\infty}^\infty xe^{-b^2x^2}\text{erf}^2(a(x-d))\ \mathrm dx &= -\text{erf}^2(a(x-d))\left.\frac{e^{-b^2x^2}}{2b^2}\right|_{-\infty}^\infty\\ &\qquad\qquad +\int_{-\infty}^\infty2\text{erf}(a(x-d))\frac{2}{\sqrt{\pi}} e^{-a^2(x-d)^2}\frac{e^{-b^2x^2}}{2b^2}\ \mathrm dx\\ &=\frac{2}{b^2\sqrt{\pi}}\int_{-\infty}^\infty\text{erf}(a(x-d)) e^{-a^2(x-d)^2-b^2x^2}\ \mathrm dx\\ &=\frac{2}{b^2\sqrt{\pi}}\int_{-\infty}^\infty\text{erf}(a(x-d)) e^{-(a^2+b^2)x^2 + 2a^2dx-a^2x^2}\ \mathrm dx \end{align*}$$ to which, after completing the square in the exponent, we can apply the OP's given integral formula $$\int_{-\infty}^{\infty}\exp\left(-b^{2}\left(x-c\right)^{2}\right)\mathrm{erf}\left(a\left(x-d\right)\right)\,\mathrm{d}x= {\frac{\sqrt\pi}{b}}\mathrm{erf}\left(\frac{ab\left(c-d\right)}{\sqrt{a^{2}+b^{2}}}\right)\,.$$

Answer 2

thanks for all your comments and suggestions. I still haven’t found the closed form of the integral, or maby it cannot be done…

What I have for this moment is a (very?) good approximation. I haven’t tested it carefully but it seems to be good enough for my purposes.

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

$$ \frac{c\sqrt{\pi}}{b}+\frac{2a}{b^{2}\sqrt{a^{2}+b^{2}}}\exp\left(-\frac{a^{2}b^{2}(c-d)^{2}}{a^{2}+b^{2}}\right)\mathrm{erf}\left(\frac{ab^{2}(c-d)}{\sqrt{a^{2}+b^{2}}\sqrt{2a^{2}+b^{2}}}\right)-\frac{c}{\sqrt{\frac{b^{2}}{\pi}+\frac{a^{2}\pi}{8}}}\exp\left(-\frac{a^{2}b^{2}(c-d)^{2}\pi^{2}}{8b^{2}+a^{2}\pi^{2}}\right) $$

for $b>0$.

Answer 3

What I managed to achieve now is a better approximation of that complicated integral, but even though I tried from various angles, I can’t find the closed form.

So basically I’m stuck with integral $$ \int_{-\infty}^{\infty}\exp\left(-a^{2}\left(x-d\right)^{2}\right)\mathrm{erf}\left(b\left(x-c\right)\right)\mathrm{erf}\left(a\left(x-d\right)\right)\,\mathrm{d}x, \tag{1} $$

which after integrating by parts leaves me with $$ \int_{-\infty}^{\infty}\exp\left(-a^{2}(x-d)^{2}\right)\mathrm{erf^{2}}\left(b\left(x-c\right)\right)\,\mathrm{d}x\,. \tag{2} $$

I’m writing this post in hope that somebody knows a trick or technique to deal with integral $(1)$ or $(2)$. Or maybe none of these integrals are “doable”?

Any suggestions will be deeply appreciated.

Notes

In the original post I mentioned one helpful integral. Another integral that is useful here is

$$ \int_{-\infty}^{\infty}\exp\left(-b^{2}\left(x-c\right)^{2}\right)\mathrm{erf}\left(a\left(x-d\right)\right)\,\mathrm{d}x= {\frac{\sqrt\pi}{b}}\mathrm{erf}\left(\frac{ab\left(c-d\right)}{\sqrt{a^{2}+b^{2}}}\right)\,. $$