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How I find this integral :

$$\int\limits_0^{\infty} x \tanh^{-1}(x)\ e^{-a x^2}\ dx$$ where $a>1$

I post same similar question but I don't know how I evaluate the above integral.

At first I use integral by part but I find divergent integral.

And also I use series same problems divergent series.

I think use for $x>1$ $\operatorname{arctanh}x=\ln\left(\frac{1+y}{y-1}\right)+ \pi i$

But problem in this ?

$$\int\limits_0^1 x \ln(1-x)e^{-ax^{2}}\ dx$$

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    $\begingroup$ Mathematica gives: $\frac{\pi e^{-a} \left(\text{erfi}\left(\sqrt{a}\right)-i\right)}{4 a}$ for $a>0$. $\endgroup$ – David G. Stork Jul 24 at 19:07
  • $\begingroup$ MMA gives this here $$\frac{e^{-a} \left(-\text{Hypergeometric1F1}^{(1,0,0)}\left(0,\frac{1}{2},a\right)-\text{Chi}(a )-\text{Shi}(a)+\log (a)+\gamma \right)}{4 a}$$ $\endgroup$ – Dr. Sonnhard Graubner Jul 24 at 19:25
  • $\begingroup$ So this question is the same as your previous question, but with $\operatorname{arctanh}(x^2)$ replaced by $\operatorname{arctanh}(x)$? $\endgroup$ – projectilemotion Jul 24 at 20:49
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As $\tanh^{-1}x=\frac{1}{2}\ln\frac{1+x}{1-x}$ for $|x|<1$, you must have $\tanh^{-1}x=\frac{1}{2}\ln\frac{x+1}{x-1}+(n-\color{red}{\frac{1}{2}})\pi i$ for some $n\in\mathbb{Z}$ when $x>1$. Making a definite choice $n\equiv 0$, we arrive at the result given by David G. Stork in his comment. Indeed, the given integral is (then) equal to $$\int_{0}^{\infty}\frac{x}{2}\ln\left|\frac{1+x}{1-x}\right|e^{-ax^2}\,dx-\frac{\pi i}{2}\int_{1}^{\infty}xe^{-ax^2}\,dx=\frac{e^{-a}}{4a}\big(I(a)-\pi i\big),$$ where \begin{align}I(a)&=\int_{0}^{\infty}\ln\left|\frac{1+x}{1-x}\right|\cdot 2axe^{a(1-x^2)}\,dx\\&=-\left.\ln\left|\frac{1+x}{1-x}\right|(e^{a(1-x^2)}\color{red}{-1})\right|_{0}^{\infty}+2\int_{0}^{\infty}\frac{e^{a(1-x^2)}-1}{1-x^2}\,dx\\&=2\int_{0}^{\infty}\frac{e^{a(1-x^2)}-1}{1-x^2}\,dx\end{align} and $I'(a)=2\int_{0}^{\infty}e^{a(1-x^2)}\,dx=e^a\sqrt{\pi/a}$. As $I(0)=0$, we get $$I(a)=\sqrt{\pi}\int_0^a\frac{e^x}{\sqrt{x}}\,dx=2\sqrt{\pi}\int_{0}^{\sqrt{a}}e^{y^2}\,dy=\pi\operatorname{erfi}\sqrt{a}.$$

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  • $\begingroup$ Thank you , but I think this divergent integral ! $\int_{0}^{\infty}\frac{e^{a(1-x^2)}-1}{1-x^2}\,dx$ $\endgroup$ – Thê Kîng Jul 24 at 21:53
  • $\begingroup$ No, it converges. The integrand is $\mathcal{O}(x^{-2})$ when $x\to\infty$, and has a removable singularity at $x=1$ (this is the whole matter of introducing that $-1$ when integrating by parts). $\endgroup$ – metamorphy Jul 25 at 3:04
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Here is my solution that is slightly different from that of metamorphy: As metamorphy explained in his answer, we are looking to find the value of $$V(a)=\int_{0}^{\infty}\frac{x}{2}\ln\left|\frac{1+x}{1-x}\right|e^{-ax^2}\,dx-\frac{\pi i}{2}\int_{1}^{\infty}xe^{-ax^2}\,dx =: J(a)-\frac{\pi i}{2}\int_{1}^{\infty}xe^{-ax^2}\,dx=J(a) - \frac{e^{-a}}{4a}\pi i.$$ To find $$J(a)=\int_0^\infty xe^{-ax^2} \left\{\frac{1}{2}\ln\left|\frac{1+x}{1-x}\right|\right\}dx$$ we integrate by parts, whereby we use $\frac{d}{dx}\left\{\frac{1}{2}\ln\left|\frac{1+x}{1-x}\right|\right\} = \frac{1}{1-x^2}$ to arrive at $$J(a)=\left. \left\{\frac{1}{2}\ln\left|\frac{1+x}{1-x}\right|\right\}\frac{-1}{2a}e^{-ax^2} \right|_0^\infty + \frac{1}{2a} \int_0^\infty \frac{e^{-ax^2}}{1-x^2}dx=\frac{1}{2a} \int_0^\infty \frac{e^{-ax^2}}{1-x^2}dx=:\frac{1}{2a}Y(a).$$ We can differentiate $$Y(a)=\int_0^\infty \frac{e^{-ax^2}}{1-x^2}dx$$ with respect to $a$ to obtain $$\begin{align}Y'(a)&=\int_0^\infty \frac{(-x^2)e^{-ax^2}}{1-x^2}dx = \int_0^\infty \frac{e^{-ax^2}}{1-x^2}(1-x^2-1)dx\\ &=\int_0^\infty e^{-ax^2}dx - \int_0^\infty \frac{e^{-ax^2}}{1-x^2}dx = \frac{1}{2}\sqrt{\frac{\pi}{a}} - Y(a)\end{align}.$$ This ordinary differential equation $$Y'(a) = \frac{1}{2}\sqrt{\frac{\pi}{a}} - Y(a)$$ can be solved by the variation of parameters: $$Y(a)=C_0 e^{-a}+ e^{-a} \int da ~\frac{1}{2}\sqrt{\frac{\pi}{a}}e^a = C_0 e^{-a}+ \frac{\pi}{2}e^{-a} \textrm{erfi}(\sqrt{a})$$ Since $Y(a=0)=0$ (using the Cauchy principal value) we have $C_0=0$ and hence $$V(a)=\frac{\pi e^{-a}}{4a}\big(\textrm{erfi}(\sqrt{a})-i\big)$$

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