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 Jul 6 answered Definite integral of product of exponential function and trigonometry function. Jun 23 revised Evaluating $\int_{-\infty}^{\infty}x\exp\left(-b^{2}\left(x-c\right)^{2}\right)\mathrm{erf}^{2}\left(a\left(x-d\right)\right)\,\mathrm{d}x$ deleted 4 characters in body Jun 23 awarded Teacher Mar 3 awarded Commentator Mar 3 comment Erf squared approximation Thanks a lot @Claude Leibovici for valuable insight into the approximation. I will clarify though that I'm not interested in the best approximation possible. In fact I need to use it to evaluate some integrals with erf squared, and my approximation seems more convenient (you can check my older posts). And BTW, could you explain what REIS is? Mar 3 awarded Curious Mar 2 comment Erf squared approximation Thanks Jack D'Aurizio. I need to digest your answer first but for now I can say that I know the approximation $\operatorname{erf}(x)^2 \approx 1-e^{-4x^2/\pi}$ but my is better. FYI the least-squares method gives $\rho^{2}=1.239$, but I find $\rho^{2}=\pi^{2}/8=1.2337$ more elegant, if I might say so :-) Mar 2 asked Erf squared approximation Sep 30 comment Integral of the product of squared exponential and two erf functions I'm not sure if you noticed my comment under your original post so I put it here once again because I'm really curious about your field of research and where that "lovely" integral appeared :-) Sep 29 answered Integral of the product of squared exponential and two erf functions Sep 24 awarded Autobiographer 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 tags Apr 27 revised Integral of product of exponential function and two complementary error functions (erfc) edited tags Apr 27 revised Differentiation under integral sign edited tags Apr 22 revised Integral of product of exponential function and two complementary error functions (erfc) added 660 characters in body Apr 21 comment Differentiation under integral sign @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. Apr 21 asked Differentiation under integral sign Apr 15 revised Integral of product of exponential function and two complementary error functions (erfc) added 664 characters in body Oct 3 awarded Fanatic Jul 25 awarded Enthusiast