11
$\begingroup$

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?

$\endgroup$
  • 3
    $\begingroup$ 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. $\endgroup$ – Omran Kouba Nov 10 '14 at 22:00
  • $\begingroup$ @OmranKouba Oh, that's really interesting! Thank you for the paper. $\endgroup$ – user 1357113 Nov 10 '14 at 22:05
  • $\begingroup$ @OmranKouba I feel the need to mention that the Example 3.2 in the paper is exceptionally nice. $\endgroup$ – user 1357113 Nov 10 '14 at 22:30
  • $\begingroup$ Yes, thank you, I think so too. $\endgroup$ – Omran Kouba Nov 11 '14 at 0:19
7
$\begingroup$

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.

$\endgroup$
12
$\begingroup$

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,

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

$ $

$(1)$ Simple Integral $\int_0^\infty (1-x\cot^{-1} x)dx=\frac{\pi}{4}$.

$(2)$ 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 $b$.

$\endgroup$
  • $\begingroup$ Good job! (+1) :-) $\endgroup$ – user 1357113 Jul 1 '15 at 18:03

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.