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mamadreqju@gmail.com


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
Apr
27
awarded  Student
Apr
27
asked 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 $