# Evaluating $\int_0^{\infty} \log(\sin^2(x))\left(1-x\operatorname{arccot}(x)\right) \ dx$

One of the ways to compute the integral

$$\int_0^{\infty} \log(\sin^2(x))\left(1-x\operatorname{arccot}(x)\right) \ dx=\frac{\pi}{4}\left(\operatorname{Li_3}(e^{-2})+2\operatorname{Li_2}(e^{-2})-2\log(2)-\zeta(3)\right)$$

is to make use of the series of $\log(\sin(x))$, but the result I got after doing that wasn't that friendly.
Is it possible to find a neat way of evaluating the integral?

• Noting that the Fourier transform of $x\mapsto(1-x\, {\rm arccot}( x))$ is given by $\omega\mapsto\dfrac{1-(1+|\omega|)e^{-|\omega|}}{\omega^2}$ the result follows from what I called "The mixed Parceval-Plancherel Formula", in arxiv.org/abs/1312.0464. This will be published in The Journal of Classical Analysis. Nov 10, 2014 at 22:00
• @OmranKouba Oh, that's really interesting! Thank you for the paper. Nov 10, 2014 at 22:05
• @OmranKouba I feel the need to mention that the Example 3.2 in the paper is exceptionally nice. Nov 10, 2014 at 22:30
• Yes, thank you, I think so too. Nov 11, 2014 at 0:19

First notice that

\begin{align} &\frac{1}{2} \int_{0}^{\infty} \log (4 \sin^{2} x) \Big(1 - x \, \text{arccot}(x) \Big) \, dx \\ &= \frac{\log(4)}{2} \int_{0}^{\infty} \Big( 1- x \, \text{arccot}(x) \Big) \, dx + \frac{1}{2} \int_{0}^{\infty} \log (\sin^{2} x) \Big(1 - x \, \text{arccot}(x) \Big) \, dx \\ &= \frac{\pi \log(2)}{4} + \frac{1}{2} \int_{0}^{\infty} \log (\sin^{2} x) \Big(1 - x \, \text{arccot}(x) \Big) \, dx . \tag{1} \end{align}

Now use the fact $$\text{Re} \log(1-e^{2ix}) = \frac{1}{2} \log(4 \sin^{2} x)$$

and integrate by parts to get

\begin{align} &\frac{1}{2} \int_{0}^{\infty} \log (4\sin^{2} x) \Big(1 - x \, \text{arccot}(x) \Big) \, dx \\ &= \text{Re} \int_{0}^{\infty} \log (1-e^{2ix}) \Big(1- x \, \text{arccot}(x) \Big) \, dx \\ &= \text{Re} \, \Big(1- x \, \text{arccot}(x) \Big) \frac{i \, \text{Li}_{2}(e^{2ix})}{2} \Bigg|^{\infty}_{0}- \text{Re} \, \frac{i}{2} \int_{0}^{\infty} \left(\frac{x}{1+x^{2}} - \text{arccot}(x) \right) \text{Li}_{2} (e^{2ix}) \, dx \\ &= 0 + \frac{1}{2} \int_{0}^{\infty} \Big(\frac{x}{1+x^{2}} - \text{arccot}(x) \Big) \sum_{n=1}^{\infty} \frac{\sin({\color{red}{2}}nx)}{n^{2}} \, dx \\ &= \frac{1}{2} \sum_{n=1}^{\infty} \frac{1}{n^{2}} \int_{0}^{\infty} \Big(\frac{x}{1+x^{2}} - \text{arccot}(x) \Big) \sin (2nx) \, dx \\ &= \frac{1}{2} \sum_{n=1}^{\infty} \frac{1}{n^{2}} \left(-\frac{\pi}{4n} + \frac{1}{n} \int_{0}^{\infty} \frac{\cos(2nx)}{(1+x^{2})^{2}} \, dx \right) \tag{2} \\ &= \frac{1}{2} \sum_{n=1}^{\infty} \frac{1}{n^{2}} \Big(-\frac{\pi}{4n} + \frac{1}{n} \frac{\pi}{4} e^{-2n} (2n+1) \Big) \tag{3} \\ &= - \frac{\pi}{8} \sum_{n=1}^{\infty} \frac{1}{n^{3}} + \frac{\pi}{4} \sum_{n=1}^{\infty} \frac{e^{-2n}}{n^{2}} + \frac{\pi}{8} \sum_{n=1}^{\infty} \frac{e^{-2n}}{n^{3}} \\ &= -\frac{\pi}{8} \zeta(3) + \frac{\pi}{4} \text{Li}_{2}(e^{-2}) + \frac{\pi}{8} \text{Li}_{3} (e^{-2}). \end{align}

Therefore,

$$\int_{0}^{\infty} \log (\sin^{2} x) \Big(1 - x \, \text{arccot}(x) \Big) \, dx = \frac{\pi}{4} \Big(\text{Li}_{3} (e^{-2}) + 2 \text{Li}_{2}(e^{-2}) -2 \log(2) - \zeta(3) \Big).$$



$(2)$ Integrate by parts again.

$(3)$ There is probably a question on here about evaluating $\int_{0}^{\infty} \frac{\cos(ax)}{(1+x^{2})^{2}} \, dx$, but I can't find it at the moment. The most direct approach is to use the residue theorem. You could also use the fact that $\int_{0}^{\infty} \frac{\cos (ax)}{b^{2}+x^{2}} \, dx = \frac{\pi}{2b} e^{-ab} \, , \, (a \ge 0,b > 0)$ and differentiate both sides with respect to $b$.

• Good job! (+1) :-) Jul 1, 2015 at 18:03

We start with: $\displaystyle \int_0^1 \frac{t^2}{x^2+t^2}\,dt = 1-x\tan^{-1}\frac{1}{x}$

Then,

\displaystyle \begin{align} \int_0^{\infty} \log (2\sin x)\left(1-x\tan^{-1}\frac{1}{x}\right)\,dx &= \int_0^{\infty}\int_0^1 \frac{t^2\log (2\sin x)}{t^2+x^2}\,dt\,dx\\&= -\sum\limits_{n=1}^{\infty} \int_0^1 t^2\int_0^{\infty} \frac{1}{n}\frac{\cos 2nx}{t^2+x^2}\,dx\,dt\tag{1}\\&= -\frac{\pi}{2}\sum\limits_{n=1}^{\infty} \frac{1}{n}\int_0^1 t^2\frac{e^{-2nt}}{t}\,dt\\&= -\frac{\pi}{2}\sum\limits_{n=1}^{\infty} \frac{1}{n}\left(\frac{1}{4n^2} - \frac{e^{-2n}}{4n^2} - \frac{e^{-2n}}{2n}\right)\\&= -\frac{\pi}{2}\left(\frac{1}{4}\operatorname{Li}_3(1)-\frac{1}{4}\operatorname{Li}_3(e^{-2})-\frac{1}{2}\operatorname{Li}_2(e^{-2})\right)\end{align}

where, we used: $\displaystyle \int_0^{\infty} \frac{\cos ax}{b^2+x^2}\,dx = \frac{\pi e^{-ab}}{2b}$ in $(1)$.

Combining with the fact that: $\displaystyle \int_0^{\infty} \left(1-x\tan^{-1}\frac{1}{x}\right)\,dx = \frac{\pi}{4}$ we get the desired result.