Evaluate the following integral : $\int\limits_0^{\infty}x\operatorname{arctanh}(x)e^{-ax^2}dx$ 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 ? 
$$\int\limits_0^1 x \ln(1-x)e^{-ax^{2}}\ dx$$
 A: 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 $$I(a)=\sqrt{\pi}\int_0^a\frac{e^x}{\sqrt{x}}\,dx=2\sqrt{\pi}\int_{0}^{\sqrt{a}}e^{y^2}\,dy=\pi\operatorname{erfi}\sqrt{a}.$$
A: 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
$$V(a)=\frac{\pi e^{-a}}{4a}\big(\textrm{erfi}(\sqrt{a})-i\big)$$
