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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