Any thoughts or hints on solving the following integral
$$ \int_{y}^{+\infty} \frac{ e^{-\sqrt{v}} }{1+ s v^{-1} } dv $$
and where $$s= \frac{2}{x^{-1}+y^{-1}}$$
The result should be a function of $x$ and $y$. Not sure if integration by parts would help.
If anyone has any approximation to the above or any way to solve it using for example any software please let me know.
I have tried the following change of variable $$u=\sqrt{v}$$ then the integral can be re-written as $$\int_{\sqrt{y}} ^{+\infty} \frac{2ue^{-u}}{1+su^{-2}} du$$ and then take perform for example an integration by part
$$x=\frac{2u}{1+su^{-2}} \ \ \ \ dx= \frac{2}{(1+su^{-2})^2}+\frac{2su^{-2}(1+2u^{-2})}{(1+su^{-2})^2} du $$ $$dy=e^{-u}du \ \ \ \ \ y= - e^{-u}$$ Then the integration can be re-written as $$xy- \int y dx$$
Thanks
Wolfram Alpha gives a closed form in terms of the exponential integral function $$Ei(x) := \int_x^\infty t^{-1} e^{-t} dt$$
See here: http://www.wolframalpha.com/input/?i=integrate+2+x%5E3+e%5E%28-x%29+%2F+%28x%5E2+%2B+s%29+dx