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 Jun 10 answered Integral of product of exponential function and two complementary error functions (erfc) Jun 9 revised Integral of product of exponential function and two complementary error functions (erfc) edited title Jun 9 asked Integral of product of exponential function and two complementary error functions (erfc) May 28 revised Evaluating $\int_{-\infty}^{\infty}x\exp\left(-b^{2}\left(x-c\right)^{2}\right)\mathrm{erf}^{2}\left(a\left(x-d\right)\right)\,\mathrm{d}x$ added 167 characters in body May 3 comment Evaluating $\int_{-\infty}^{\infty}x\exp\left(-b^{2}\left(x-c\right)^{2}\right)\mathrm{erf}^{2}\left(a\left(x-d\right)\right)\,\mathrm{d}x$ Thanks Dilip. I did calculations you suggested but unfortunately I can’t see how this result is supposed to help me. With $c\neq 0$ I still have integrand $\exp\left(-a^2 (-d + x)^2\right) \mathrm{erf}\left(b (x-c)\right) \mathrm{erf}\left(a (x-d)\right)$ and I don’t know how to proceed with it. May 3 answered Evaluating $\int_{-\infty}^{\infty}x\exp\left(-b^{2}\left(x-c\right)^{2}\right)\mathrm{erf}^{2}\left(a\left(x-d\right)\right)\,\mathrm{d}x$ Apr 29 awarded Supporter Apr 29 accepted Integral with exp and erf Apr 29 revised Integral with exp and erf corrected the right side of the integral, it’s now sqrt(pi)/b instead of sqrt(pi/b) Apr 29 comment Integral with exp and erf Thanks a lot! It explains all I wanted to know. And yes, I made a typo in the first post: it should be $\frac{\sqrt{\pi}}{b}$. Apr 29 revised Evaluating $\int_{-\infty}^{\infty}x\exp\left(-b^{2}\left(x-c\right)^{2}\right)\mathrm{erf}^{2}\left(a\left(x-d\right)\right)\,\mathrm{d}x$ added 2 characters in body Apr 29 awarded Scholar Apr 29 answered Evaluating $\int_{-\infty}^{\infty}x\exp\left(-b^{2}\left(x-c\right)^{2}\right)\mathrm{erf}^{2}\left(a\left(x-d\right)\right)\,\mathrm{d}x$ Apr 29 revised Integral with exp and erf added 6 characters in body Apr 28 asked Integral with exp and erf Apr 27 comment Evaluating $\int_{-\infty}^{\infty}x\exp\left(-b^{2}\left(x-c\right)^{2}\right)\mathrm{erf}^{2}\left(a\left(x-d\right)\right)\,\mathrm{d}x$ @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. Apr 27 revised Evaluating $\int_{-\infty}^{\infty}x\exp\left(-b^{2}\left(x-c\right)^{2}\right)\mathrm{erf}^{2}\left(a\left(x-d\right)\right)\,\mathrm{d}x$ edited body Apr 27 awarded Editor Apr 27 comment Evaluating $\int_{-\infty}^{\infty}x\exp\left(-b^{2}\left(x-c\right)^{2}\right)\mathrm{erf}^{2}\left(a\left(x-d\right)\right)\,\mathrm{d}x$ @AdamRubinson I meant $[erf(a(x-d))]^2$. I edited the original post to avoid confusion. Apr 27 revised Evaluating $\int_{-\infty}^{\infty}x\exp\left(-b^{2}\left(x-c\right)^{2}\right)\mathrm{erf}^{2}\left(a\left(x-d\right)\right)\,\mathrm{d}x$ edited body