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 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? 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 :-) 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 :-) 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. May3 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. Apr29 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 $\frac{\sqrt{\pi}}{b}$. Apr27 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. Apr27 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.