I want to find an elementary evaluation of

$$I=\int_0^\infty \left(\frac{\sqrt\pi}2\operatorname{erfi}(x)e^{-x^2}-\frac1{1+2x}\right)dx$$ where $\operatorname{erfi}(x)=\frac{2}{\sqrt\pi}\int_0^xe^{t^2}dt$.

Rough Solution
$$I=\int_0^\infty\left({}_1F_1(1;3/2,-x^2)x-\frac1{1+2x}\right)dx$$ $$=\left(\frac{x^2}2{}_2F_2(1,1;3/2,2,-x^2)-\frac12\ln(1+2x)\right)\Bigg|_0^\infty$$ By using the asymptotic expansion of $_2F_2$ I can get the answer is $\frac{\gamma}4$, where $\gamma$ is the Euler's constant.
I wonder if there is a elementary proof without using hypergeometric function.

  • 2
    $\begingroup$ $erfi(x)$ and $erf(x)$ , which one is yours? In general $\operatorname{erfi}(x)=\frac{2}{\sqrt{\pi}}\int_0^xe^{z^2}dz,$ and $\operatorname{erf}(x)=\frac{2}{\sqrt{\pi}}\int_0^xe^{-z^2}dz.$ $\endgroup$ – Riemann Sep 14 '18 at 9:25
  • $\begingroup$ @Riemann. According to the result, it is $\text{erfi}$ $\endgroup$ – Claude Leibovici Sep 14 '18 at 9:57
  • 1
    $\begingroup$ Probably not useful, but $$I=\int_0^\infty\,\left(\frac{\sqrt{\pi}}{2}\,\text{erfc}(x)\,\exp\left(+x^2\right)-\frac{1}{1+2x}\right)\,\text{d}x\,.$$ Here, $\text{erfc}(x)=1-\text{erf}(x)$ for all $x\in\mathbb{C}$. (Of course, I assume that $\text{erf}$ and $\text{erfi}$ are defined correctly as Riemann suggested.) $\endgroup$ – Batominovski Sep 14 '18 at 10:19
  • $\begingroup$ @KemonoChen Can you please clarify your question. $\endgroup$ – paulplusx Sep 14 '18 at 13:16
  • $\begingroup$ It is $\operatorname{erfi}$. Sorry for the typo. $\endgroup$ – Kemono Chen Sep 14 '18 at 13:16

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