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 ?

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

  • 1
    $\begingroup$ 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

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 $$I(a)=\sqrt{\pi}\int_0^a\frac{e^x}{\sqrt{x}}\,dx=2\sqrt{\pi}\int_{0}^{\sqrt{a}}e^{y^2}\,dy=\pi\operatorname{erfi}\sqrt{a}.$$

  • $\begingroup$ 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

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 $$V(a)=\frac{\pi e^{-a}}{4a}\big(\textrm{erfi}(\sqrt{a})-i\big)$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.