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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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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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  • $\begingroup$ Thanks, Robert. I've continued from your suggestion below. $\endgroup$ – Donagh Dec 10 '12 at 12:12
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Let us define: \begin{eqnarray} J(\alpha)&:=& \int\limits_0^\infty erf\left( \frac{\alpha}{\sqrt{1+x}} - \frac{\sqrt{1+x}}{\beta} \right) e^{-\frac{x}{\gamma}} dx\\ &\underbrace{=}_{t=\sqrt{1+x}}& \int\limits_1^\infty erf\left( \frac{\alpha}{t} - \frac{t}{\beta} \right) e^{-\frac{t^2-1}{\gamma}} 2 t dt \end{eqnarray} Now we differentiate with respect to $\alpha$. We have: \begin{eqnarray} &&\frac{\partial}{\partial \alpha} J(\alpha)=\\ && \int\limits_1^\infty \frac{4}{\sqrt{\pi}} e^{-(\frac{\alpha}{t} - \frac{t}{\beta})^2} e^{-\frac{t^2-1}{\gamma}} dt =\\ && \frac{4}{\sqrt{\pi}} e^{2 \frac{\alpha}{\beta}+\frac{1}{\gamma}} \int\limits_1^\infty e^{-\left( \frac{\alpha^2}{t^2} + (\frac{1}{\beta^2}+\frac{1}{\gamma}) t^2\right)} dt \underbrace{=}_{\begin{array}{rrr} A&:=&\alpha^2\\B&:=&1/\beta^2+1/\gamma\end{array}}\\ && \frac{4}{\sqrt{\pi}} e^{2 \frac{\alpha}{\beta}+\frac{1}{\gamma}} \cdot \left.\frac{\sqrt{\pi } \left(e^{-2 \sqrt{A} \sqrt{B}} \left(1-\text{erf}\left(\frac{\sqrt{A}}{t}-\sqrt{B} t\right)\right)+e^{2 \sqrt{A} \sqrt{B}} \left(\text{erf}\left(\frac{\sqrt{A}}{t}+\sqrt{B} t\right)-1\right)\right)}{4 \sqrt{B}} \right|_1^\infty =\\ % &&= \frac{e^{\frac{1}{g}}}{\sqrt{\frac{1}{b^2}+\frac{1}{g}}} \left(e^{a \left(-\Delta_-\right)} \text{erf}\left(a-\sqrt{\frac{1}{b^2}+\frac{1}{g}}\right)-e^{a \left(\Delta_+\right)} \text{erf}\left(a+\sqrt{\frac{1}{b^2}+\frac{1}{g}}\right)+e^{a \left(-\Delta_-\right)}+e^{a \left(\Delta_+\right)}\right) \end{eqnarray} where $\delta:=\sqrt{\frac{1}{b^2}+\frac{1}{g}}$ and $\Delta_\pm:=2 \delta \pm \frac{2}{b} $.

Now all we need to do is to integrate over $\alpha$. A glimpse at the bottom line above suffices to realize that the anti-derivative with respect to $\alpha$ can be easily found using integration by parts. In addition it is also easy to see that $J(\infty) = \gamma$. Therefore the final result reads: \begin{eqnarray} \gamma- J(\alpha) = % \gamma\cdot (1-erf(\alpha-\frac{1}{\beta})) + \frac{\gamma e^{\frac{1}{\gamma}} e^{-\alpha \Delta_-}}{2 \delta} \left(\frac{1}{\beta} - \frac{\Delta_-}{2} e^{4 \alpha \delta} +\delta\right) + \frac{\gamma e^{\frac{1}{\gamma}}}{4 \delta} \left( \Delta_+ e^{-\alpha \Delta_-} erf(a-\delta) + e^{\alpha \Delta_+} \Delta_- erf(a+\delta) \right) \end{eqnarray}

In[441]:= {b, g} = RandomReal[{0, 2}, 2, WorkingPrecision -> 50];
dd = Sqrt[1/b^2 + 1/g];
{Dm, Dp} = 2 {-1/b + dd, 1/b + dd};
a = Range[0, 2, 0.001];
f = Interpolation[
  Transpose[{a, 
    NIntegrate[
       Erf[#/Sqrt[1 + x] - Sqrt[1 + x]/b] Exp[-x/g], {x, 0, 
        Infinity}] & /@ a}]]



In[446]:= a = RandomReal[{0, 2}, WorkingPrecision -> 50];
f'[a]
E^(1/g)/Sqrt[
 1/b^2 + 1/
  g] (E^(a (2 /b - 2 Sqrt[1/b^2 + 1/g])) + E^(
   a (2 /b + 2  Sqrt[1/b^2 + 1/g])) + 
   E^(a (2 /b - 2  Sqrt[1/b^2 + 1/g])) Erf[a - Sqrt[1/b^2 + 1/g]] - 
   E^(a (2 /b + 2 Sqrt[1/b^2 + 1/g])) Erf[a + Sqrt[1/b^2 + 1/g]])

g - f[a]


 g (1 - Erf[a - 1/b]) + (g E^(1/g) E^(-a Dm))/(
  2 dd) (1/b + E^(4 a Sqrt[1/b^2 + 1/g]) (-Dm/2) + dd) + (g E^(1/g))/(
  4 dd) ((Dp) E^(-a Dm) Erf[a - dd] + E^(a Dp) (Dm) Erf[a + dd])


Out[447]= 0.648692

Out[448]= 0.648691540832040074541903643440002542062490436

Out[449]= 0.681556

Out[450]= 0.681555757815934967025234698639289684849178686
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