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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 $b>0$.

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    Do you mean erf((a(x-d))^2), or [erf(a(x-d))]^2 (in the top expression)?2012-04-27
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    @AdamRubinson I meant $[erf(a(x-d))]^2$. I edited the original post to avoid confusion.2012-04-27
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    In that case, I am guessing that they gave you the integral at the bottom of your post as a hint. Note that, assuming your hint at the bottom is correct, that integral is just a constant (even though it looks incredibly messy). You can use IBP where one of your functions is the integrand of the integral at the bottom of your post, and the other function is erf(a(x-d)). The "hint" at the bottom of your post tells you how to integrate the longer one, and all you need to do now is to differentiate erf(a(x-d)) (which you can do)2012-04-27
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    @AdamRubinson Well, after the initial excitement I realized that it’s not that simple. I cannot use the bottom integral because what I need is the antiderivative of $x\exp\left(-b^{2}\left(x-c\right)^{2}\right)\mathrm{erf}\left(a\left(x-d\right)\right)$, and not the value of the integral.2012-04-27
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    yes you are correct. My method fails. This is not an easy problem, and I don't know the answer.2012-04-27

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