How to express an integral involving an erf,rational function and a gaussian in terms of special functions?

Question

I am trying to find a version of the following integral in terms of special functions:

$$ \int _{\alpha }^{\beta }\frac{e^{-r^2 \sigma ^2} \text{erf}(\gamma \sigma )}{\sigma }d\sigma \quad (1) $$ for parameters $r>0, \alpha>0, \beta>0, \gamma>0.$ $\beta = \infty$ possible.

I have tried integration by parts, with $\frac{e^{-r^2 \sigma ^2} }{\sigma }$ being the function to be integrated, which then leads me to an integral involving the exponential integral and a Gaussian: $$ \int_\alpha^\beta\operatorname{Ei}\left(-\sigma^2\right)e^{-\mu \sigma^2}d\sigma $$ for $\mu = \frac{\gamma}{r}.$ But this didn't take me further.

I also tried taking the derivative with respect to $\gamma$ and then integrating for $\sigma$ like here, leaving me with: $$ \int_0^\gamma\frac{\text{erf}\left(\sigma \sqrt{\tau ^2+r^2}\right)}{\sqrt{\tau^2+r^2}} d\tau $$

Does anyone know if this (1) or any of the other integrals can be simplified or expressed in terms of special functions? Any help would be greatly appreciated!

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Special Case: I found the result for $\alpha = 0, \beta = \infty:$ $$ \int _0^{\infty }\frac{e^{-r^2 \sigma ^2} \text{erf}(\gamma \sigma )}{\sigma }d\sigma = \sinh ^{-1}\left(\frac{\gamma }{r}\right) $$ which can be achieved by integration by parts, and using the known integral $$ \begin{equation} \int_0^\infty\operatorname{Ei}\left(-x^2\right)e^{-\mu x^2}dx=-\sqrt{\frac{\pi}{\mu}}\operatorname{arcsinh}\sqrt{\mu} \end{equation} $$ for $\mu = \frac{\gamma}{r}.$ But I´m more interested in the case where $\alpha, \beta>0$ are any positive values.

Answer 1

$$I=\int_0^\gamma\frac{\text{erf}\left(\sigma \sqrt{\tau ^2+r^2}\right)}{\sqrt{\tau^2+r^2}}\, d\tau$$ Writing $$\frac{\text{erf}\left(\sigma \sqrt{\tau ^2+r^2}\right)}{\sqrt{\tau^2+r^2}}=\frac{2 \sigma }{\sqrt{\pi }}\sum_{n=0}^\infty (-1)^n\,\frac{ \left(r^2+\tau ^2\right)^n}{(2 n+1)\, n!}\,\sigma ^{2 n}$$ then $$I=\frac{2 \,\gamma\, \sigma }{\sqrt{\pi }}\,\,\sum_{n=0}^\infty (-1)^n\,\frac{(r \sigma )^{2 n}}{(2 n+1)\, n!}\,\,\, _2F_1\left(\frac{1}{2},-n;\frac{3}{2};-\frac{\gamma ^2}{r^2}\right)$$ where $\,_2F_1\left(\frac{1}{2},-n;\frac{3}{2 };-x\right)$ is a simple polynomial of degree $n$ in $x$.

A few numerical tests seem to be correct.

Answer 2

$$\int_{\alpha }^{\beta } \frac{\exp \left(-r^2 x^2\right) \text{erf}(\gamma x)}{x} \, dx=\\\int_{\alpha }^{\beta } \frac{\left(\sum _{n=0}^{\infty } \frac{\left(-r^2 x^2\right)^n}{n!}\right) \sum _{m=0}^{\infty } \frac{2 (-1)^m (x \gamma )^{1+2 m}}{\sqrt{\pi } (m!+2 m m!)}}{x} \, dx=\\\sum _{n=0}^{\infty } \sum _{m=0}^{\infty } \int_{\alpha }^{\beta } \frac{\left(-r^2 x^2\right)^n \left(2 (-1)^m (x \gamma )^{1+2 m}\right)}{(n! x) \left(\sqrt{\pi } (m!+2 m m!)\right)} \, dx=\\\sum _{m=0}^{\infty } \sum _{n=0}^{\infty } \frac{2 (-1)^{m+n} r^{2 n} \left(-\alpha ^{1+2 m+2 n}+\beta ^{1+2 m+2 n}\right) \gamma ^{1+2 m}}{(1+2 m) (1+2 m+2 n) \sqrt{\pi } m! n!}=\\\sum _{m=0}^{\infty } \sum _{n=0}^{\infty } -\frac{(2 \alpha \gamma ) \left(\left(\frac{1}{2}\right)_m \left(\frac{1}{2}\right)_{m+n}\right) \left(\left(-r^2 \alpha ^2\right)^n \left(-\alpha ^2 \gamma ^2\right)^m\right)}{\sqrt{\pi } \left(\left(\frac{3}{2}\right)_m \left(\frac{3}{2}\right)_{m+n}\right) (n! m!)}+\sum _{m=0}^{\infty } \sum _{n=0}^{\infty } \frac{(2 \beta \gamma ) \left(\left(\frac{1}{2}\right)_m \left(\frac{1}{2}\right)_{m+n}\right) \left(\left(-r^2 \beta ^2\right)^n \left(-\beta ^2 \gamma ^2\right)^m\right)}{\sqrt{\pi } \left(\left(\frac{3}{2}\right)_m \left(\frac{3}{2}\right)_{m+n}\right) (n! m!)}$$

Two last double infinty series can be expressed by:Kampé de Fériet function.

also:$$\int_{\alpha }^{\beta } \frac{\exp \left(-r^2 x^2\right) \text{erf}(\gamma x)}{x} \, dx=\sum _{m=0}^{\infty } \frac{(-1)^m \left(\frac{\gamma }{r}\right)^{1+2 m} \left(\Gamma \left(\frac{1}{2}+m,r^2 \alpha ^2\right)-\Gamma \left(\frac{1}{2}+m,r^2 \beta ^2\right)\right)}{(1+2 m) \sqrt{\pi } m!}$$

Answer 3

I prefer to add a second answer for the initial problem $$I=\int _{\alpha }^{\beta }\frac{e^{-r^2 \sigma ^2} \text{erf}(\gamma \sigma )}{\sigma }\,d\sigma=\int _{0 }^{\beta }\frac{e^{-r^2 \sigma ^2} \text{erf}(\gamma \sigma )}{\sigma }\,d\sigma-\int ^{\alpha }_{0 } \frac{e^{-r^2 \sigma ^2} \text{erf}(\gamma \sigma )}{\sigma }\,d\sigma $$ Using again the series expansion of the error function $$\frac{e^{-r^2 \sigma ^2} \text{erf}(\gamma \sigma )}{\sigma }=\frac{2}{\sqrt{\pi }}\sum_{n=0}^\infty (-1)^n\,\frac {\gamma ^{2 n+1} }{(2 n+1)\, n! }\,\,\sigma ^{2 n}\,\, e^{-r^2 \sigma ^2}$$

$$J_n=\int_0^t \sigma ^{2 n}\,\, e^{-r^2 \sigma ^2}\,d\sigma=\frac{1 }{2\, r^{2 n+1} }\left(\Gamma \left(n+\frac{1}{2}\right)-\Gamma \left(n+\frac{1}{2},r^2 t^2\right)\right)$$ So, if $$K(t)=\int _0^t\frac{e^{-r^2 \sigma ^2} \text{erf}(\gamma \sigma )}{\sigma }\,d\sigma$$ $$\color{blue}{K(t)=\sinh ^{-1}\left(\frac{\gamma }{r}\right)-\frac{1}{\sqrt{\pi }}\sum_{n=0}^\infty (-1)^n\,\frac{\left(\frac{\gamma }{r}\right)^{2 n+1}}{(2n+1)\, n!}\,\Gamma \left(n+\frac{1}{2},r^2 t^2\right)}$$

If $a_n$ is the summand $$\underset{n\to \infty }{\text{limit }}\left|\frac{a_{n+1}}{a_n}\right|=\left(\frac{\gamma }{r}\right)^2$$

Edit

Considering $$I=\int_0^t\operatorname{Ei}\left(-\sigma^2\right)e^{-\mu \sigma^2}\,d\sigma$$ using $$\operatorname{Ei}\left(-\sigma^2\right)=2 \log (\sigma )+\gamma+\sum_{n=1}^\infty (-1)^n\, \frac{\sigma^{2n}}{n\,n! }$$ $$I=\frac{\sqrt{\pi }\, (2 \log (t)+\gamma )\, \text{erf}\left(\sqrt{\mu } t\right)}{2 \sqrt{\mu }}-2 \,t \,\, _2F_2\left(\frac{1}{2},\frac{1}{2};\frac{3}{2}, \frac{3}{2};-\mu t^2 \right)-$$ $$ \sqrt{\frac{\pi}{\mu }}\log \left(\frac{1}{2} \left(1+\sqrt{1+\frac{1}{\mu }}\right)\right)-\frac 12\sum_{n=1}^\infty (-1)^n\,\frac {\Gamma \left(n+\frac{1}{2},\mu t^2 \right) } {n\, n!\, \mu ^{n+\frac{1}{2}} }$$