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I'd like to calculate the following integral:

$$\int^{\infty}_{0} \mathrm{erf}\left(\frac{\alpha}{\sqrt{1+x}} - \frac{\sqrt{1+x}}{\beta}\right) \exp\left(-\frac{x}{\gamma}\right)\, dx,$$

where $\beta > 0$, $\gamma > 0$ and $\alpha \in \mathbb{R}$.

I've tried a few approaches, but with no success.

The form is similar to Equation 12 on page 177 of Erdelyi's Tables of Integral Transforms (Vol. 1):

$$\int^{\infty}_{0} \mathrm{erf}\left(\frac{\alpha}{\sqrt{t}} - \frac{\sqrt{t}}{\beta}\right) \exp\left(-\frac{t}{\gamma}\right)\, dt$$

but the change of variables requires a change in limits.

Any advice would be greatly appreciated!

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up vote 2 down vote accepted

If you let: $$t = \sqrt{1 + x},$$

and change limits appropriately, Mathematica evaluates it as:

$$\gamma \text{erf}\left(\alpha -\frac{1}{\beta }\right) ++\frac{\gamma ^{3/2} e^{\frac{2 \alpha }{\beta }-2 |\alpha | \sqrt{\frac{1}{\beta ^2}+\frac{1}{\gamma }}+\frac{1}{\gamma }} \left(|\alpha | \left(\text{erf}\left(\sqrt{\frac{1}{\beta ^2}+\frac{1}{\gamma }}-|\alpha |\right)-e^{4 |\alpha | \sqrt{\frac{1}{\beta ^2}+\frac{1}{\gamma }}} \text{erfc}\left(|\alpha |+\sqrt{\frac{1}{\beta ^2}+\frac{1}{\gamma }}\right)-1\right)+\alpha \sqrt{\frac{\beta ^2+\gamma }{\gamma }} \left(\text{erf}\left(\frac{\sqrt{\frac{\beta ^2+\gamma }{\gamma }}}{\beta }-|\alpha |\right)+e^{\frac{4 |\alpha | \sqrt{\frac{\beta ^2+\gamma }{\gamma }}}{\beta }} \text{erfc}\left(|\alpha |+\frac{\sqrt{\frac{\beta ^2+\gamma }{\gamma }}}{\beta }\right)-1\right)\right)}{2 |\alpha | \sqrt{\beta ^2+\gamma }}.$$

Not as neat as I'd like it, but there you go.

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I'd start with an integration by parts, which should give you $$ \text{erf}(\alpha - 1/\beta) \gamma - \frac{\gamma}{\beta \sqrt{\pi}} \exp(2\alpha/\beta - 1/\beta^2) \int_0^\infty \exp\left(-(1/\beta^2+1/\gamma) x - \alpha^2/(1+x) \right) \frac{\alpha \beta + 1 + x}{(1+x)^{3/2}} \ dx$$ Now according to Maple, for $A > 0$ $$ \int_0^\infty \exp(-A x - \alpha^2/(1+x)) \dfrac{dx}{(1+x)^{1/2}} = 1/2\,{\frac {{{\rm e}^{A-2\,{\alpha}\,\sqrt {A}}} \left( {{\rm erf}\left(-\sqrt {A}+{\alpha}\right)}+1+ \left( 1- {{\rm erf}\left(\sqrt {A}+{\alpha}\right)} \right) {{\rm e}^{4\,{ \alpha}\,\sqrt {A}}} \right) \sqrt {\pi }}{\sqrt {A}}}$$ However, I wasn't able to get a closed form for the integral with $(1+x)^{3/2}$ instead of $(1+x)^{1/2}$.

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Thanks, Robert. I've continued from your suggestion below. – Donagh Dec 10 '12 at 12:12

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