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I know how to solve $\int_{a}^{b}\frac{1}{x}e^{-x^{2}}dx$ using upper incomplete gamma function (we can assum that $\left|b\right|>\left|a\right|$) $$\int_{a}^{b}\frac{1}{x}e^{-x^{2}}dx=\int_{a}^{b}\frac{2x}{2x^{2}}e^{-x^{2}}dx\stackrel{\begin{array}{c} t=x^{2}\\ dt=2xdx \end{array}}{=}\int_{a^{2}}^{b^{2}}\frac{1}{2t}e^{-t}dt=\frac{1}{2}\left[\Gamma\left(0,a^{2}\right)-\Gamma\left(0,b^{2}\right)\right]$$ there is something that I can do when I have some phase in the exponent: $$\int_{a}^{b}\frac{1}{x}e^{-\left(x+c\right)^{2}}dx=?$$ Even an evaluation will help,

thanks!

uzi harush
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    There really seems to be no closed form for the phased version, but there may be a series evaluation or a version in terms of a more general function. – Тyma Gaidash Dec 04 '21 at 15:57
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    There may be a closed form. It will not be neat though. Let $I(\alpha)=\int_a^b dx \ e^{-(x+c)^2+\alpha x}$, which may be evaluated in terms of the error function. The requested integral is $\int d\alpha \ I(\alpha) $, and this integral is in the form of Gaussian times error function which you may find here and here – Sal Dec 07 '21 at 19:51
  • @Sal Maybe you can post a solution or a start of a solution as an answer? – Тyma Gaidash Dec 07 '21 at 20:21
  • @TymaGaidash I will if I make any progress. My first attempt resulted in a divergent integral though... – Sal Dec 07 '21 at 22:38

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