# Is there a closed form for $\int_{-\infty}^{\infty} \frac{e^{-ax^2}}{\mathrm{erfc}{(-bx)}} dx$?

The integral expression is $$I = \int_{-\infty}^{\infty} \frac{e^{-ax^2}}{\mathrm{erfc}{(-bx)}} dx$$ where $a>0$ and $b>0$.

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Please clarify what you are asking. –  rschwieb May 4 '12 at 1:02
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## 1 Answer

This integral most likely does not admit a closed form in terms of known special functions.

Using $$\begin{eqnarray} \int_{-\infty}^{\infty} \mathrm{e}^{-a x^2} \frac{ \mathrm{d} x}{\operatorname{erfc}(-b x)} &=& \int_{-\infty}^{\infty} \mathrm{e}^{(b^2-a) x^2} \mathrm{d} \left(\log (\operatorname{erfc}(-b x)) \right) \\ &=& \left. \mathrm{e}^{(b^2-a) x^2} \log (\operatorname{erfc}(-b x)) \right|_{x \downarrow -\infty}^{x \uparrow +\infty} - \int_{-\infty}^\infty 2 x (b^2-a) \mathrm{e}^{(b^2-a) x^2} \log (\operatorname{erfc}(-b x)) \mathrm{d} x \end{eqnarray}$$ Since $\log (\operatorname{erfc}(-b x))$ is at most polynomial growth, it follows that convergence requires $a > b^2$.

Indeed, here is a plot of the integrand for $b=1$ and $a = \frac{1}{2}$:

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