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 Mar3 awarded Commentator Mar3 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? Mar3 awarded Curious Mar2 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 :-) Mar2 asked Erf squared approximation Sep30 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 :-) Sep29 answered Integral of the product of squared exponential and two erf functions Sep24 awarded Autobiographer Apr27 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 Apr27 revised Integral of product of exponential function and two complementary error functions (erfc) edited tags Apr27 revised Differentiation under integral sign edited tags Apr22 revised Integral of product of exponential function and two complementary error functions (erfc) added 660 characters in body Apr21 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. Apr21 asked Differentiation under integral sign Apr15 revised Integral of product of exponential function and two complementary error functions (erfc) added 664 characters in body Oct3 awarded Fanatic Jul25 awarded Enthusiast Jun10 answered Integral of product of exponential function and two complementary error functions (erfc) Jun9 revised Integral of product of exponential function and two complementary error functions (erfc) edited title Jun9 asked Integral of product of exponential function and two complementary error functions (erfc)