Find the integral in a closed form : $\int\limits_0^\infty\text{arctanh}(x^{2})e^{-ax^{2}}dx$

Question

Evaluate :

$\displaystyle\int_{0}^{\infty}x\text{arctanh}(x^{2})e^{-ax^{2}}dx$ where $a>1$

I don't know if we can find closed form or not but I have some results for $a=1,2,3..$

$\displaystyle\int_{0}^{\infty}x\text{arctanh}(x^{2})e^{-x^{2}}dx=\frac{π\text{erfi}(1)}{e}$

$\displaystyle\int_{0}^{\infty}x\text{arctanh}(x^{2})e^{-2x^{2}}dx=\frac{π\text{erfi}(\sqrt {2})}{2e^{2}}$

$\displaystyle\int_{0}^{\infty}x\text{arctanh}(x^{2})e^{-3x^{2}}dx=\frac{π\text{erfi}(\sqrt {3})}{3e^{3}}$

So I think the closed form is :

$\displaystyle\int_{0}^{\infty}x\text{arctanh}(x^{2})e^{-ax^{2}}dx=\frac{π\text{erfi}(\sqrt {a})}{ae^{a}}$

But how can I prove it ?

I know $\text{arctanh}(x)=\sum_{n=0}^{\infty}\frac{x^{2n+1}}{2n+1}$

But when I use it, I find the series to be divergent!

Answer 1

We need to find:

$$I=\frac{1}{2} \int_0^\infty \tanh^{-1} (y) e^{-a y} dy=\frac{1}{2} \int_0^1 \tanh^{-1} (y) e^{-a y} dy+\frac{1}{2} \int_1^\infty \tanh^{-1} (y) e^{-a y} dy$$

For $y<1$ we have:

$$\tanh^{-1} (y)=\frac{1}{2} \left(\log(1+y)-\log(1-y)\right)$$

For $y>1$ we have:

$$\tanh^{-1} (y)=\frac{1}{2} \left(\log(y+1)-\log(y-1)- \pi i \right)$$

So the imaginary part of the integral is equal to:

$$\Im I=\frac{\pi }{4} \int_1^\infty e^{-a y} dy=\frac{\pi e^{-a} }{4a} $$

The real part consists of three different integrals:

$$\Re I=\frac{1}{4} (I_1+I_2+I_3)$$

$$I_1=\int_0^\infty \log(1+y) e^{-a y} dy=-\frac{e^a}{a} \operatorname{Ei}(-a)$$

Where we have the exponential integral.

$$I_2=-\int_0^1 \log(1-y) e^{-a y} dy=-\frac{e^{-a}}{a} \left(\gamma+\log a-\operatorname{Ei}(a) \right)$$

$$I_3=-\int_1^\infty \log(y-1) e^{-a y} dy=\frac{e^{-a}}{a}(\gamma+\log a)$$

This simplifies to:

$$I_2+I_3=\frac{e^{-a}}{a} \operatorname{Ei}(a)$$

So we finally have:

$$\Re I=-\frac{1}{4a} \left(e^a \operatorname{Ei}(-a)-e^{-a} \operatorname{Ei}(a) \right)$$

$$\Im I=\frac{\pi e^{-a} }{4a}$$

Let's check numerically. For $a=3$ we have from Wolfram Alpha:

$$2I=0.1261100236\ldots -0.0260684480380\ldots i$$

This is the same as the values given by the closed forms above.

The $\operatorname{erfi}$ result from the OP is not correct.