In my research appeared the following integral:

$$\int e^{-a x^2 } \frac {b x^2}{c + k x} dx$$

Where $a$, $b$, $c$ and $k$ are constants.

I know that the result of this indefinite integral is no conventional function and that such an integral can be solved numerically.

I tried it in all sites for online calculation of integrals with no results.

Has anyone seen this integral before? Is there any special function, like erf or gamma, that was defined to express this integral?

If not possible, perhaps there is some general expression for the integral defined from $0$ to $\infty$

PS: The integral above arose from the product of a function with the Gaussian probability function

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    $\begingroup$ For an integral to be improper it must have, at least, limits... $\endgroup$ – DonAntonio Apr 26 '19 at 22:39
  • $\begingroup$ It can be expressed in terms of the incomplete gamma function. $\endgroup$ – Count Iblis Apr 26 '19 at 22:42
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    $\begingroup$ I think you might get $\pm \infty$ across $x=-\dfrac{c}{k}$. Perhaps easier to see if you use the substitution $y=c+kx$ $\endgroup$ – Henry Apr 26 '19 at 22:44
  • $\begingroup$ You are right, @quasi, I got confused. Now the constants are with the proper names $\endgroup$ – Paulo Buchsbaum Apr 26 '19 at 22:45
  • $\begingroup$ How is it possible, @CountIblis? $\endgroup$ – Paulo Buchsbaum Apr 26 '19 at 22:47

If you make $c+k x=t$ and expand, you end with $$I=\frac{b c^2 }{k^3 }\int\frac{ e^{-\frac{a (t-c)^2}{k^2}}}{ t}\,dt-\frac{2 b c }{k^3}\int e^{-\frac{a (t-c)^2}{k^2}}\,dt+\frac{b }{k^3}\int t e^{-\frac{a (t-c)^2}{k^2}}\,dt$$ The problem is with the first integral (the second and third do not make any problem).

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  • $\begingroup$ It's a big progress. I'm trying to find something about first part of the sum. If one replaces $x = (t-c)/k$ back in first part, one gets $ \frac{bc^2}{k^2} \int e^{- a x^2}/(x-c)$. $\endgroup$ – Paulo Buchsbaum Apr 27 '19 at 4:42
  • $\begingroup$ However, I can't find nothing about the first part, in the original way written in your answer or in the changed first part. in the above comment. I've tried all online indefinite integral calculators, incluiding Wolfram Alpha, integral-calculator.com, symbolab.com, emathhelp.net, etc. , neither using exponential integral (en.wikipedia.org/wiki/Exponential_integral) $\endgroup$ – Paulo Buchsbaum Apr 27 '19 at 5:28
  • $\begingroup$ Errata: The first part is $ (bc^2/k) \int e^{-ax^2}/(x-c) dx$ because if one replaces $x= (t-c)/k$ then $dx = k dt$ $\endgroup$ – Paulo Buchsbaum Apr 27 '19 at 12:57

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