# Closed form for the integral $\int_0^1 \frac{dy}{1+y^2} \tanh^{-1} \frac{1}{\sqrt{3+y^2}}$

This integral (in a slightly different form) appeared in the book of Borwein and Devlin with a remark that no closed form exists for it so far:

$$I=\int_0^1 \frac{dy}{1+y^2} \tanh^{-1} \frac{1}{\sqrt{3+y^2}}$$

It was even named among "impossible integrals" (page 58).

I have spent several hours on this integral and found no closed form of course, but I don't think it's hopeless.

With the most obvious series expansion and some work (and liberal use of Mathematica), I have obtained the following explicit series for the integral:

$$I= \log(1+\sqrt{2})\arctan \frac{1}{\sqrt{2}}- \frac{1}{6} \sum_{n=0}^\infty \frac{1}{(2n+1)2^n}\sum_{l=0}^{n-1} \left(\frac{2}{3} \right)^l \sum_{k=0}^{l} \frac{(-1)^k}{(2k+1)4^k} \binom{l}{k}$$

The series has good convergence, by the way:

Is there a more simple series expression for the integral, or, who knows, maybe even a closed form was found after the book was published?

## Too long for comment...

Using the Euler substitution $\sqrt{3+x^{2}}=x+t$, we find that the integral transforms as

\begin{align} \mathcal{I} &=\int_{0}^{1}\mathrm{d}x\,\frac{\operatorname{arctanh}{\left(\frac{1}{\sqrt{3+x^{2}}}\right)}}{1+x^{2}}\\ &=\int_{\sqrt{3}}^{1}\mathrm{d}t\,\frac{\left(-1\right)\left(3+t^{2}\right)}{2t^{2}}\cdot\frac{1}{1+\left(\frac{3-t^{2}}{2t}\right)^{2}}\\ &~~~~~\times\operatorname{arctanh}{\left(\frac{2t}{3+t^{2}}\right)};~~~\small{\left[\sqrt{3+x^{2}}=x+t\right]}\\ &=\int_{1}^{\sqrt{3}}\mathrm{d}t\,\frac{2\left(t^{2}+3\right)}{\left(t^{4}-2t^{2}+9\right)}\operatorname{arctanh}{\left(\frac{2t}{3+t^{2}}\right)}\\ &=\sqrt{3}\int_{\frac{1}{\sqrt{3}}}^{1}\mathrm{d}u\,\frac{2\left(3u^{2}+3\right)}{\left(9u^{4}-6u^{2}+9\right)}\operatorname{arctanh}{\left(\frac{2\sqrt{3}\,u}{3+3u^{2}}\right)};~~~\small{\left[t=\sqrt{3}\,u\right]}\\ &=\frac{2}{\sqrt{3}}\int_{\frac{1}{\sqrt{3}}}^{1}\mathrm{d}u\,\frac{\left(u^{2}+1\right)}{\left(u^{4}-\frac23u^{2}+1\right)}\operatorname{arctanh}{\left(\frac{1}{\sqrt{3}}\cdot\frac{2u}{1+u^{2}}\right)}.\\ \end{align}

Setting $\alpha:=\arcsin{\left(\frac{1}{\sqrt{3}}\right)}\in\left(0,\frac{\pi}{4}\right)$, note that

$$\cos{\left(2\alpha\right)}=1-2\sin^{2}{\left(\alpha\right)}=\frac13.$$

Hence,

\begin{align} \mathcal{I} &=\frac{2}{\sqrt{3}}\int_{\frac{1}{\sqrt{3}}}^{1}\mathrm{d}u\,\frac{\left(u^{2}+1\right)}{\left(u^{4}-\frac23u^{2}+1\right)}\operatorname{arctanh}{\left(\frac{1}{\sqrt{3}}\cdot\frac{2u}{1+u^{2}}\right)}\\ &=2\sin{\left(\alpha\right)}\int_{\sin{\left(\alpha\right)}}^{1}\mathrm{d}u\,\frac{u^{2}+1}{u^{4}-2u^{2}\cos{\left(2\alpha\right)}+1}\operatorname{arctanh}{\left(\frac{2u\sin{\left(\alpha\right)}}{1+u^{2}}\right)}\\ &=\sin{\left(\alpha\right)}\int_{\sin{\left(\alpha\right)}}^{1}\mathrm{d}u\,\frac{u^{2}+1}{u^{4}-2u^{2}\cos{\left(2\alpha\right)}+1}\ln{\left(\frac{1+\left(\frac{2u\sin{\left(\alpha\right)}}{u^{2}+1}\right)}{1-\left(\frac{2u\sin{\left(\alpha\right)}}{u^{2}+1}\right)}\right)}\\ &=\sin{\left(\alpha\right)}\int_{\sin{\left(\alpha\right)}}^{1}\mathrm{d}u\,\frac{u^{2}+1}{\left[u^{2}-2u\cos{\left(\alpha\right)}+1\right]\left[u^{2}+2u\cos{\left(\alpha\right)}+1\right]}\\ &~~~~~\times\ln{\left(\frac{u^{2}+2u\sin{\left(\alpha\right)}+1}{u^{2}-2u\sin{\left(\alpha\right)}+1}\right)}\\ &=\frac12\sin{\left(\alpha\right)}\int_{\sin{\left(\alpha\right)}}^{1}\mathrm{d}u\,\left[\frac{1}{u^{2}+2u\cos{\left(\alpha\right)}+1}+\frac{1}{u^{2}-2u\cos{\left(\alpha\right)}+1}\right]\\ &~~~~~\times\left[\ln{\left(u^{2}+2u\sin{\left(\alpha\right)}+1\right)}-\ln{\left(u^{2}-2u\sin{\left(\alpha\right)}+1\right)}\right].\\ \end{align}

Thus, the integral $\mathcal{I}$ can be reduced to a sum of integrals of the form,

$$\int\mathrm{d}x\,\frac{\ln{\left(x+p\right)}}{x+q};~~~\small{\left(p,q\right)\in\mathbb{C}^{2}},$$

so in principle, it should be possible to evaluate $\mathcal{I}$ in terms of dilogarithms with complex arguments...