# Integrating a product of exponential and complementary error function with square-root of variable in the denominator

I need to evaluate \begin{equation} \int_a^\infty \mathrm{erfc}\left( \frac{b}{\sqrt{c\cdot h}} \right) e^{-d\cdot h} dh \end{equation} where $\mathrm{erfc}(s) = \frac{2}{\sqrt{\pi}} \int_{s}^{\infty} \exp(-t^2) dt$.

A closed-form expression is appreciated since ultimately, I need to do

\begin{equation} \int_0^\infty \left( \int_{k\cdot y}^\infty \mathrm{erfc}\left( \frac{b}{\sqrt{c\cdot h}} \right) e^{-d\cdot h} dh \right) e^{-m \cdot y} dy \end{equation}

I've noticed that a similar function - the Q-function - such that \begin{align} Q(s) &= \frac{1}{\sqrt{2\pi}} \int_s^\infty e^{-\frac{x^2}{2}}dx \\ &=\frac{1}{2} \mathrm{erfc}(\frac{x}{\sqrt{2}}) \end{align} and the Q-function has an alternative representation \begin{align} Q(s) = \frac{1}{\pi} \int_0^{\frac{\pi}{2}} \exp{\left(\frac{-s^2}{2\sin^2{\phi}} \right)}d\phi \end{align} but I'm not sure if this helps.

Assume $b,c,d\neq0$ and $d>0$ to avoid the trivial cases or the divergent cases:

$\int_a^\infty\mathrm{erfc}\biggl(\dfrac{b}{\sqrt{ch}}\biggr)e^{-dh}~dh$

$=-\int_a^\infty\mathrm{erfc}\biggl(\dfrac{b}{\sqrt{ch}}\biggr)d\biggl(\dfrac{e^{-dh}}{d}\biggr)$

$=-\biggl[\dfrac{e^{-dh}}{d}\mathrm{erfc}\biggl(\dfrac{b}{\sqrt{ch}}\biggr)\biggr]_a^\infty+\int_a^\infty\dfrac{e^{-dh}}{d}d\biggl(\mathrm{erfc}\biggl(\dfrac{b}{\sqrt{ch}}\biggr)\biggr)$

$=\dfrac{e^{-ad}}{d}\mathrm{erfc}\biggl(\dfrac{b}{\sqrt{ac}}\biggr)+\dfrac{b}{d\sqrt c}\int_\infty^ae^{-\frac{b^2}{ch}-dh}~d\biggl(\dfrac{1}{\sqrt{h}}\biggr)$

$=\dfrac{e^{-ad}}{d}\mathrm{erfc}\biggl(\dfrac{b}{\sqrt{ac}}\biggr)+\dfrac{b}{d\sqrt c}\int_0^\frac{1}{\sqrt{a}}e^{-\frac{b^2h^2}{c}-\frac{d}{h^2}}~dh$

$=\dfrac{e^{-ad}}{d}\mathrm{erfc}\biggl(\dfrac{b}{\sqrt{ac}}\biggr)-\biggl[\dfrac{b\sqrt\pi}{4|b|d}\biggl(e^\frac{2|b|\sqrt d}{\sqrt c}~\mathrm{erfc}\biggl(\dfrac{|b|h}{\sqrt c}+\dfrac{\sqrt d}{h}\biggr)+e^{-\frac{2|b|\sqrt d}{\sqrt c}}\mathrm{erfc}\biggl(\dfrac{|b|h}{\sqrt c}-\dfrac{\sqrt d}{h}\biggr)\biggr)\biggr]_0^\frac{1}{\sqrt{a}}$

(using the result in http://dlmf.nist.gov/7.7#E7)

$=\dfrac{e^{-ad}}{d}\mathrm{erfc}\biggl(\dfrac{b}{\sqrt{ac}}\biggr)+\dfrac{b\sqrt\pi}{4|b|d}\biggl(e^{-\frac{2|b|\sqrt d}{\sqrt c}}\mathrm{erfc}\biggl(\sqrt a\sqrt d-\dfrac{|b|}{\sqrt a\sqrt c}\biggr)-e^\frac{2|b|\sqrt d}{\sqrt c}~\mathrm{erfc}\biggl(\sqrt a\sqrt d+\dfrac{|b|}{\sqrt a\sqrt c}\biggr)\biggr)$