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Any idea how to perform these two integrations?

1. $$ \int_{0}^{\infty} \frac{exp(-a x^2)}{x^2(x^2+\kappa^2)} dx $$ 2. $$ \int_{0}^{\infty} \frac{exp(-a x^2)}{(x^2+\kappa^2)} dx $$ second integration is same as equation 1.42 of this link

isoura
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1 Answers1

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I am afraid that these integrals do not show explicit expression.

Considering the second one $$I=\int \frac{e^{-a x^2}}{x^2+\kappa ^2}\,dx$$ let $x=\kappa y$ and $b=a \kappa ^2$ to make $$I=\frac 1 \kappa \int \frac{e^{-b y^2}}{y^2+1}\,dy=\frac1 \kappa\int e^{-by^2}\Big[\sum_{n=0}^\infty (-1)^n y^{2n}\Big]\,dy$$ $$I=\frac1 \kappa\sum_{n=0}^\infty (-1)^n\int y^{2n}\, e^{-b y^2}\,dy=\frac1 \kappa\int e^{-b y^2}\,dy+\frac1 \kappa\sum_{n=1}^\infty (-1)^n\int y^{2n}\, e^{-b y^2}\,dy$$ with $$\int e^{-b y^2}\,dy=\frac{\sqrt{\pi } }{2 \sqrt{b}}\,\text{erf}\left(\sqrt{b} y\right)$$ $$\int y^{2n}\, e^{-b y^2}\,dy=b^{-(n+\frac{1}{2})} \Gamma \left(n+\frac{1}{2},b y^2\right)$$

  • that's very useful; I was trying to get the answer as given in eq 1.42, in the following link,http://www.mhtlab.uwaterloo.ca/courses/me755/web_chap2.pdf, I have updated the question, as I found this later – isoura Mar 30 '20 at 05:22
  • @isoura. Your question was not about the definite integral but about th antiderivative. Please clarify. – Claude Leibovici Mar 30 '20 at 05:42
  • Yeah, it is for the antiderivative only, but I'm trying to get the definite integral as well. – isoura Mar 30 '20 at 08:36