Closed form for ${\large\int}_0^\infty\frac{\arctan(x)\,\operatorname{arccot}(x+1)}{x}dx$

Question

I'm looking for a closed form for this integral: $$I=\int_0^\infty\frac{\arctan(x)\,\operatorname{arccot}(x+1)}{x}dx.$$ Mathematica and Maple could not evaluate it symbolically. Numerically, $$I\approx1.3513049368715095284050230093075694014884142059538...$$ WolframAlpha and ISC+ could not find a plausible closed form for this number. Still, I hope that it exists, because the integral looks nice.

Answer 1

The integrand has a closed-form antiderivative in terms of elementary functions and polylogarithms. It can be found using Mathematica after expressing inverse trig functions through logarithms of complex arguments, and can be manually checked for correctness using differentiation. After subtracting its limits at $\infty$ and $0$ and simplification, we can get this result: $$\begin{align}I&=\frac1{32}\Big[\!\operatorname{Li}_2\!\left(\tfrac15\right)\cdot\ln2+\operatorname{Li}_3\!\left(\tfrac15\right)+ \operatorname{Li}_3\!\left(\tfrac45\right)\!\Big]+\frac{\pi}{16}\,\left(3\operatorname{Ti}_2(2)-4\,G\right)\\&+\frac{\ln5}{192}\,\left(9\ln2\cdot\ln5-12\ln^22-2\ln^25\right)+\frac{\pi^2}{192}\,\left(\ln5-7\ln2\right)+\frac{17}{16}\zeta(3),\end{align}$$ where $\operatorname{Ti}_2(x)$ is the inverse tangent integral, $G=\operatorname{Ti}_2(1)$ is the Catalan constant.

Answer 2

Using Mathematica I have arrived at the result

$$I=\frac{1}{4} (-\text{Li}_3(-1-i)-\text{Li}_3(-1+i)+\text{Li}_3(1-i)+\text{Li}_3(1+i))+\frac{1}{4} \left(-\text{Li}_3\left(-\frac{1}{2}-\frac{i}{2}\right)-\text{Li}_3\left(-\frac{1}{2}+\frac{i}{2}\right)+\text{Li}_3\left(\frac{1}{2}-\frac{i}{2}\right)+\text{Li}_3\left(\frac{1}{2}+\frac{i }{2}\right)+\text{Li}_2\left(-\frac{1}{2}-\frac{i}{2}\right) \log \left(-\frac{1}{2}-\frac{i}{2}\right)+\text{Li}_2\left(-\frac{1}{2}+\frac{i}{2}\right) \log \left(-\frac{1}{2}+\frac{i}{2}\right)-\text{Li}_2\left(\frac{1}{2}-\frac{i}{2}\right) \log \left(\frac{1}{2}-\frac{i}{2}\right)-\text{Li}_2\left(\frac{1}{2}+\frac{i}{2}\right) \log \left(\frac{1}{2}+\frac{i}{2}\right)\right)$$

Which matches your numerical estimates.

$$I\approx1.3513049368715095284050230093075694014884142059538$$

I am unable to simplify it past the polylogs. Running full simplify we can get to

$$\frac{1}{384} \left(48 \pi C+3 \left(-64 \text{Li}_3\left(-\frac{1}{2}-\frac{i}{2}\right)-64 \text{Li}_3\left(-\frac{1}{2}+\frac{i}{2}\right)+32 (\text{Li}_3(1-i)+\text{Li}_3(1+i))+35 \zeta (3)\right)+\text{Li}_2\left(-\frac{1}{2}-\frac{i}{2}\right) (-48 \log (2)-72 i \pi )+24 \text{Li}_2\left(-\frac{1}{2}+\frac{i}{2}\right) (-\log (4)+3 i \pi )-4 \log ^3(2)+7 \pi ^2 \log (2)\right)$$ Where $C$ is Catalan's constant.

I think this is about as closed form as it is going to get.