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While working on mixture (variance) of normal distribution and keep running into these two integrals

$ \int_0^{\infty}\dfrac{v}{\sqrt{v + c}}e^{-\dfrac{y^2}{2(v + c)} - \dfrac{(u-v)^2}{u^2v}}dv,$

$\int_0^{\infty}\dfrac{v^{-1}}{\sqrt{v + c}}e^{-\dfrac{y^2}{2(v + c)} - \dfrac{(u-v)^2}{u^2v} }dv,$

where $c>0, u>0 ,y\in \mathbb R$.

I was wondering are they solvable? Can they be expressed as some known function or in elementary terms?

Any help would be appreciated.

1 Answers 1

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For this integral, the best result seems to be a Taylor series in $c$ with coefficients in closed form.

In the typical case $c=1$, even specializing to the easiest case of $u=1$, $y=0$, we get the integrals $\int_0^\infty v^{\pm1}(v+1)^{-1/2}e^{-v-1/v}dv$, for which neither Mathematica nor Gradshteyn and Ryzhik has any answer.

However, there is an explicit expression in the limit case $c=0$, which includes Gradshteyn and Ryzhik 3.471.15. Setting $z=\sqrt{4+2y^2}/u$, Mathematica gives the two integrals as: $\frac{u^3(1+z)}{2}\sqrt{π}\,e^{\,z(\sqrt{2}−1)} \text{ and } \frac{2}{zu}\sqrt{π}\,e^{\,z(\sqrt{2}−1)}.$

When we expand the integrands above as power series in $c$, the integrals of each term have similar closed-form expressions.