Can anyone solve these two integrals .

$$ \int_{0}^{ \infty } \frac{x^2 e^{-x^2/2 \sigma ^2}}{(x-a)^2+b^2} dx $$


$$ \int_{0}^{ \infty } \frac{e^{-(\ln x - \mu )^2/2 \sigma ^2}}{(x-a)^2+b^2} dx $$

Either give me some hint or tell me the procedure, if anybody knows how to solve these type of exponential integrals.



closed as off-topic by Jeel Shah, John Gowers, Ali Caglayan, N. F. Taussig, graydad Jan 14 '15 at 0:50

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  • $\begingroup$ Mathematica can find the first integral in terms of the exponential integral special function. $\endgroup$ – user111187 Jan 14 '15 at 7:57
  • $\begingroup$ it's gonna be a little bit messy to get a closed form solution, but it's in principle doable. Maybe tomorrow i find the time $\endgroup$ – tired Jan 14 '15 at 15:58
  • $\begingroup$ can you tell me which procedure to use .. in order to solve this integral $\endgroup$ – Anjan Tripathi Jan 19 '15 at 21:28

I'm going to assume that $a$ and $b$ are both real numbers and that $b \neq 0$. Then the function $\frac{z^{2}e^{-z^{2}/\sigma^{2}}}{(z-a)^{2} + b^{2}}$ for complex-valued $z$ is meromorphic with poles at $z = a \pm bi$.

Then my idea would be to try integrating over a quarter pie-slice contour: $\Gamma_{R} = [0,R] \cup i[0,R] \cup \{ Re^{i\theta} \, : \, \theta \in [0,\pi/2]\}$, oriented counter-clockwise. Then apply the Cauchy residue formula, and if one can show the integrals over the second and third components of $\Gamma_{R}$ go to zero as $R \to \infty$, we are done.

  • 2
    $\begingroup$ $$e^{-(i t)^2/\sigma^2} = e^{t^2/\sigma^2}$$ $\endgroup$ – Ron Gordon Jan 13 '15 at 21:44

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