PDF of the product of normal and Cauchy distributions I am having trouble in finding out the resulting PDF of the product of normal and Cauchy distributions. It turns out that we have a general formula for calculating the PDF of product of two random distribution but the integral is not converging . Also, I tried using Mellin Transform method but it is getting too complicated. 
If anybody has any idea about how to approach the problem, please share it with me.
Thanks

Assuming the normal is centered with variance $s$ and the Cauchy distribution has parameters $a, b$, combining this Wikipedia page and Mathematica gives
$$
\text{ConditionalExpression}\left[\frac{i \left(\frac{e^{-\frac{z^2}{2 s^2 (a-i b)^2}}
   \left(2 \pi  \text{erfi}\left(\frac{\left| z\right| }{\sqrt{2} (a s-i b s)}\right)+2
   \text{Ei}\left(\frac{z^2}{2 (a-i b)^2 s^2}\right)+\log \left(\frac{s^2 (a-i
   b)^2}{z^2}\right)-\log \left(\frac{z^2}{s^2 (a-i b)^2}\right)-4 \log (-a+i b)-2 \log
   \left(\frac{s^2}{z^2}\right)\right)}{a-i b}-\frac{e^{-\frac{z^2}{2 s^2 (a+i b)^2}}
   \left(2 \pi  \text{erfi}\left(\frac{\left| z\right| }{\sqrt{2} (a s+i b s)}\right)+2
   \text{Ei}\left(\frac{z^2}{2 (a+i b)^2 s^2}\right)-\log \left(\frac{1}{(a+i
   b)^2}\right)+4 \log \left(-\frac{1}{a+i b}\right)+\log \left((a+i
   b)^2\right)\right)}{a+i b}\right)}{8 \sqrt{2} \pi ^{3/2} s}-\frac{i
   \left(\frac{e^{-\frac{z^2}{2 s^2 (a-i b)^2}} \left(2 \pi 
   \text{erfi}\left(\frac{\left| z\right| }{\sqrt{2} (a s-i b s)}\right)-2
   \text{Ei}\left(\frac{z^2}{2 (a-i b)^2 s^2}\right)+\log \left(\frac{1}{(a-i
   b)^2}\right)+4 \log (a-i b)-\log \left((a-i b)^2\right)\right)}{a-i
   b}+\frac{e^{-\frac{z^2}{2 s^2 (a+i b)^2}} \left(-2 \pi  \text{erfi}\left(\frac{\left|
   z\right| }{\sqrt{2} (a s+i b s)}\right)+2 \text{Ei}\left(\frac{z^2}{2 (a+i b)^2
   s^2}\right)-\log \left(\frac{1}{(a+i b)^2}\right)-4 \log (a+i b)+\log \left((a+i
   b)^2\right)\right)}{a+i b}\right)}{8 \sqrt{2} \pi ^{3/2} s},b\neq 0\land a=0\right]
$$
