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

I need to evaluate $$\int_a^\infty \mathrm{erfc}\left( \frac{b}{\sqrt{c\cdot h}} \right) e^{-d\cdot h} dh$$ 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

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

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.

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## 1 Answer

Hint:

$\int\mathrm{erfc}\biggl(\dfrac{b}{\sqrt{ch}}\biggr)e^{-dh}~dh=-\dfrac{e^{-dh}}{d}\mathrm{erfc}\biggl(\dfrac{b}{\sqrt{ch}}\biggr)-\dfrac{be^{-2\sqrt{b^2}\sqrt{\frac{d}{c}}}\biggl(\mathrm{erf}\biggl(\dfrac{\sqrt{b^2}-ch\sqrt{\frac{d}{c}}}{\sqrt{ch}}\biggr)+e^{4\sqrt{b^2}\sqrt{\frac{d}{c}}}\biggl(\mathrm{erf}\biggl(\dfrac{\sqrt{b^2}+ch\sqrt{\frac{d}{c}}}{\sqrt{ch}}\biggr)-1\biggr)+1\biggr)}{2d\sqrt{b^2}}+C$

$\therefore\int_a^\infty\mathrm{erfc}\biggl(\dfrac{b}{\sqrt{ch}}\biggr)e^{-dh}~dh=\dfrac{e^{-ad}}{d}\mathrm{erfc}\biggl(\dfrac{b}{\sqrt{ac}}\biggr)+\dfrac{be^{-2\sqrt{b^2}\sqrt{\frac{d}{c}}}\biggl(\mathrm{erf}\biggl(\dfrac{\sqrt{b^2}-ac\sqrt{\frac{d}{c}}}{\sqrt{ac}}\biggr)+e^{4\sqrt{b^2}\sqrt{\frac{d}{c}}}\biggl(\mathrm{erf}\biggl(\dfrac{\sqrt{b^2}+ac\sqrt{\frac{d}{c}}}{\sqrt{ac}}\biggr)-1\biggr)+1\biggr)}{2d\sqrt{b^2}}-\lim\limits_{h\to\infty}\left(\dfrac{e^{-dh}}{d}\mathrm{erfc}\biggl(\dfrac{b}{\sqrt{ch}}\biggr)+\dfrac{be^{-2\sqrt{b^2}\sqrt{\frac{d}{c}}}\biggl(\mathrm{erf}\biggl(\dfrac{\sqrt{b^2}-ch\sqrt{\frac{d}{c}}}{\sqrt{ch}}\biggr)+e^{4\sqrt{b^2}\sqrt{\frac{d}{c}}}\biggl(\mathrm{erf}\biggl(\dfrac{\sqrt{b^2}+ch\sqrt{\frac{d}{c}}}{\sqrt{ch}}\biggr)-1\biggr)+1\biggr)}{2d\sqrt{b^2}}\right)$

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