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Before you answer this OP, please read all the terms and conditions below. Thank you...

Today I hold an unofficial little contest on brilliant.org. Now, I will hold it here on Math S.E. It's just for fun guys. (>‿◠)✌

Before we start the contest, here are the rules of my little contest that you should obey as a contestant (the one who post an answer to this OP during contest period):

  1. The contest will be started on Sunday, October $26$, $2014$ at $12.00$ p.m. noon (UTC). So, until D-Day, please do not make any attempts (posting comments or answers) that can lead users here to answer the problem contest.
  2. You may post your answer after this OP is marked as a bounty question by me. I will give a bounty to this OP at least $250$ reps. The bounty will be awarded to the winner.
  3. The winner of this contest is the highest voted answer. So, the jury of this contest is not only me, but all of us.
  4. Each contestant can only post one answer and the answer can be posted anytime during the contest period. Once the answer is posted, it cannot be edited during the contest period. Since you have at least $4$ days to prepare the answer to this OP, please make sure you create a zero error answer.
  5. Do not post this contest problem on other sites for asking a hint or an answer.
  6. The contest will be ended on Sunday, November $2$, $2014$ (after bounty period over).
  7. The contestant will be disqualified for violating the contest rules above. The term disqualified means the contestant who violates the rules but she/ he owns the highest voted answer at the end of the contest will not be declared as the winner. Therefore, the bounty will not be awarded to her/ him.

If necessary, the rules may be changed accordingly. To be fair, I will not answer this OP nor give any hints to anyone. I really need your cooperation in order to make this contest succeed. Thank you.


Here is the contest problem:

Prove\begin{equation}\large{\int_0^{\Large\frac{\pi}{2}}}\ln\left(\frac{\ln^2\sin\theta}{\pi^2+\ln^2\sin\theta}\right)\,\frac{\ln\cos\theta}{\tan\theta}\,d\theta=\frac{\pi^2}{4}\end{equation}

Feel free to use any methods to solve this problem (real or complex analysis approach). Okay, happy problem solving and best of luck! ٩(˘◡˘)۶

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    $\begingroup$ I have purged the comments to this question, which all deal with meta-issues best handled on the related meta-thread. $\endgroup$ – user642796 Oct 25 '14 at 9:18
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Note: I will be making this post Community Wiki as it bears resemblance to Chris's sis's answer.


I will use the following result: $$\lim_{N\to\infty}\left[\sum^\infty_{k=1}\frac{(-1)^{k+1}}{2k(2k)!}N^{2k}-\ln{N}\right]=\gamma$$ It's derivation can be found here.


Letting $\ln(\sin{\theta})=-x$ and using $\mathcal{I}$ to denote the integral in question, \begin{align} \mathcal{I} =&\frac{1}{2}\int^\infty_0\ln(1-e^{-2x})\ln\left(\frac{x^2}{\pi^2+x^2}\right)\ {\rm d}x\\ =&-\frac{\partial}{\partial a}\Bigg{|}_{a=0}\sum^\infty_{n=1}\frac{1}{n}\int^\infty_0x^ae^{-2nx}\ {\rm d}x+\frac{1}{2}\sum^\infty_{n=1}\frac{1}{n}\int^\infty_0e^{-2nx}\ln(\pi^2+x^2)\ {\rm d}x\\ =&-\frac{\partial}{\partial a}\Bigg{|}_{a=0}\frac{\Gamma(a+1)\zeta(a+2)}{2^{a+1}}+\frac{1}{2}\sum^\infty_{n=1}\frac{\ln{\pi}}{n^2}+\frac12\sum^\infty_{n=1}\frac{1}{n^2}\int^\infty_0\frac{xe^{-2nx}}{\pi^2+x^2}{\rm d}x\\ =&-\frac{1}{2}\Gamma'(1)\zeta(2)-\frac{1}{2}\Gamma(1)\zeta'(2)+\frac{1}{2}\Gamma(1)\zeta(2)\ln{2}+\frac{1}{2}\zeta(2)\ln{\pi}\\ &+\frac{1}{2}\sum^\infty_{n=1}\frac{1}{n}\int^\infty_0\int^\infty_0e^{-2nx}e^{-xy}\cos{\pi y}\ {\rm d}x\ {\rm d}y\\ =&\frac{\pi^2}{12}\left(\gamma+\ln{2\pi}\right)-\frac{1}{2}\zeta'(2)+\frac12\sum^\infty_{n=1}\frac{1}{n^2}\int^\infty_0\frac{\cos{\pi y}}{y+2n}{\rm d}y\\ =&\frac{\pi^2}{12}\left(\gamma+\ln{2\pi}\right)-\frac{1}{2}\zeta'(2)+\frac12\sum^\infty_{n=1}\frac{1}{n^2}\int^\infty_0\frac{\cos{(y+2n\pi)}}{y+2n\pi}{\rm d}y\\ =&\frac{\pi^2}{12}\left(\gamma+\ln{2\pi}\right)-\frac{1}{2}\zeta'(2)+\frac12\sum^\infty_{n=1}\frac{1}{n^2}\int^\infty_{2n\pi}\frac{\cos{y}}{y}{\rm d}y\\ =&\frac{\pi^2}{12}\left(\gamma+\ln{2\pi}\right)-\frac{1}{2}\zeta'(2)+\frac12\sum^\infty_{n=1}\frac{1}{n^2}\left[\left(\int^\infty_0-\int^{2n\pi}_0\right)\frac{\cos{y}-1}{y}{\rm d}y+\ln{y}\Bigg{|}^\infty_{2n\pi}\right]\\ =&\color{grey}{\frac{\pi^2}{12}\left(\gamma+\ln{2\pi}\right)-\frac{1}{2}\zeta'(2)}+\frac12\sum^\infty_{n=1}\frac{1}{n^2}\int^{2n\pi}_0\frac{1-\cos{y}}{y}{\rm d}y\color{grey}{-\frac{1}{2}\sum^\infty_{n=1}\frac{\ln(2n\pi)}{n^2}}\\ &+\color{grey}{\frac{1}{2}\lim_{N\to\infty}\sum^\infty_{n=1}\frac{1}{n^2}\left[\ln{N}-\sum^\infty_{k=1}\frac{(-1)^{k+1}}{2k(2k)!}N^{2k}\right]}\\ =&\frac{1}{2}\sum^\infty_{n=1}\frac{1}{n^2}\int^1_0\frac{1-\cos{2n\pi y}}{y}{\rm d}y\\ =&\frac{1}{2}\int^1_0\left(\frac{\pi^2}{6y}-\pi^2y+\pi^2-\frac{\pi^2}{6y}\right)\ {\rm d}y=\frac{\pi^2}{2}\int^1_0(1-y)\ {\rm d}y=\Large{\frac{\pi^2}{4}} \end{align} as was to be shown.

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    $\begingroup$ Brilliant answer as usual & more important it is easy to understand for a high school student like me, +1 ≧◠‿◠≦✌ $\endgroup$ – Anastasiya-Romanova 秀 Oct 27 '14 at 15:23
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    $\begingroup$ @Anastasiya-Romanova Easy to understand for a high school student like you maybe, but I think I speak for most of us when I say I wouldn't have understood this in high school! ^_^; $\endgroup$ – David H Oct 30 '14 at 17:47
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I am also a proponent of the opinion that the proposed rules are against the way how we communicate ideas in this community. At the same time, however, the mathematical part of OP is something worth it to be dealt with. So here is a solution:


1. Preliminary

Before the calculation we make some preliminary results:

Lemma 1. For any $u > 0$ and $n > 0$, we have $$\frac{1}{n^{2}} \log \left(1 + \frac{4\pi^{2}n^{2}}{u^{2}} \right) = \pi^{2} \int_{u/2}^{\infty} \frac{2}{s^{2} + n^{2}\pi^{2}} \, \frac{ds}{s}.$$

Proof. Differentiating both sides with respect to $u$, we check that they must equal up to a constant. Taking $u \to \infty$, we find that this constant should equal zero. ////

Lemma 2. For any real $x$, we have $$ \sum_{n=1}^{\infty} \frac{2}{s^{2} + n^{2}\pi^{2}} = \frac{s \coth s - 1}{s^{2}}. $$

Although non-trivial, this is a standard result in complex analysis. So we omit the proof.

Lemma 3. Let $f(s) = (1 - e^{-2s})(s\coth s - 1)$. Then

  • $f(s) = (s-1) + (s+1)e^{-2s}$ and hence $f''(s) = 4s e^{-2s}$.
  • $f(s)/s^{2}$ and $f'(s)/s$ converges to $0$ as $s \to 0$ and $s \to +\infty$.

Proof. The first assertion is just a simple calculation. To prove the second assertion, it suffices to look into the McLaurin series expansion $f(s) = \frac{2}{3}s^{3} - \frac{2}{3}s^{4} + \cdots$. ////

2. Calculation

Now we are ready to calculate the integral. Let $I$ denote the integral. Then with the substitution $\sin^{2}\theta = e^{-t}$ (so that $d\theta/\tan\theta = -dt/2t$), we have

\begin{align*} I &= \frac{1}{2} \int_{0}^{\frac{\pi}{2}} \log \left( \frac{\log^{2} \sin^{2}\theta}{4\pi^{2} + \log^{2}\log^{2}\theta} \right) \frac{\log^{2}\cos^{2}\theta}{\tan\theta} \, d\theta \\ &= \frac{1}{4} \int_{0}^{\infty} \log(1 - e^{-t}) \log\left( \frac{t^{2}}{4\pi^{2} + t^{2}} \right) \, dt\\ &= \frac{1}{4} \int_{0}^{\infty} \sum_{n=1}^{\infty} \frac{e^{-nt}}{n} \log \left(1 + \frac{4\pi^{2}}{t^{2}} \right) \, dt. \end{align*}

Now we utilize the Tonelli's theorem to interchange the summation and integral. Then

\begin{align*} I &= \frac{1}{4} \sum_{n=1}^{\infty} \int_{0}^{\infty} \frac{e^{-nt}}{n} \log \left(1 + \frac{4\pi^{2}}{t^{2}} \right) \, dt \\ &= \frac{1}{4} \sum_{n=1}^{\infty} \int_{0}^{\infty} \frac{e^{-u}}{n^{2}} \log \left(1 + \frac{4\pi^{2}n^{2}}{u^{2}} \right) \, du, \quad (u = nt) \\ &= \frac{\pi^{2}}{4} \sum_{n=1}^{\infty} \int_{0}^{\infty} e^{-u} \left( \int_{u/2}^{\infty} \frac{2}{s^{2} + n^{2}\pi^{2}} \, \frac{ds}{s} \right) \, du, \end{align*}

where the last equality follows from Lemma 1. Applying the Tonelli's theorem again, Lemma 2 shows that

\begin{align*} I &= \frac{\pi^{2}}{4} \int_{0}^{\infty} e^{-u} \left( \int_{u/2}^{\infty} \frac{s \coth s - 1}{s^{2}} \, \frac{ds}{s} \right) \, du \\ &= \frac{\pi^{2}}{4} \int_{0}^{\infty} \left( \int_{0}^{2s} e^{-u} \, du \right) \frac{s \coth s - 1}{s^{3}} \, ds \\ &= \frac{\pi^{2}}{4} \int_{0}^{\infty} \frac{f(s)}{s^{3}} \, ds, \end{align*}

where we applied Tonelli's theorem again in the second line, and $f(s)$ denotes the function in Lemma 3. So it suffices to prove that the last integral, without the constant $\pi^{2}/4$, equals 1. Indeed, Lemma 3 shows that

\begin{align*} \int_{0}^{\infty} \frac{f(s)}{s^{3}} \, ds &= \left[ -\frac{f(s)}{2s^{2}} \right]_{0}^{\infty} + \frac{1}{2} \int_{0}^{\infty} \frac{f'(s)}{s^{2}} \, ds \\ &= \left[ -\frac{\smash{f'}(s)}{2s} \right]_{0}^{\infty} + \frac{1}{2} \int_{0}^{\infty} \frac{f''(s)}{s} \, ds \\ &= \int_{0}^{\infty} 2e^{-2s} \, ds = 1 \end{align*}

and therefore we get $I = \pi^{2}/4$ as desired.

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    $\begingroup$ I can sincerely agree with your first statement in your answer. No worries. Anyway, it's such a great pleasure for me to get an answer of my OP from a great integrator like you (ɔ◔‿◔)ɔ ♥ +1 $\endgroup$ – Anastasiya-Romanova 秀 Oct 26 '14 at 12:18
  • $\begingroup$ @Anastasiya-Romanova, Thank you! And I just found that this approach is quite similar to that of the solution of Eric Naslund for a similar problem. $\endgroup$ – Sangchul Lee Oct 26 '14 at 12:26
  • $\begingroup$ Ya, you're right. I created this integral in my scrap-paper using several journals. I thought it would be hard to crack but I'm totally wrong. I am aware of your answer to that OP, I've upvoted too long time ago but only your answer, now I upvote Mr. @EricNaslund (I summon him to post his approach here). Oh ya, I posted this problem at Brilliant.org and Quora in case you wanna answer it there $\endgroup$ – Anastasiya-Romanova 秀 Oct 26 '14 at 12:39
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Here's a solution using complex analysis. As other people have shown, the integral is equivalent to: $$I=\frac{1}{2}\int_0^\infty\log\bigg(\frac{x^2}{x^2+\pi^2}\bigg)\log(1-e^{-2x})dx$$ I'll integrate the function: $$f(z)=\log z \log(1-e^{-2z})$$ On this contour: enter image description here

I'll spare you the gory details of showing that for $\epsilon \rightarrow 0$ and $R \rightarrow \infty$ the integrals on $\gamma_2$, $\gamma_4$ and $\gamma_6$ vanish. We are left with: $$\int_0^\infty \log x\log(1-e^{-2x})dx +\int_\infty^0\log(x+i\pi)\log(1-e^{-2x})dx+i\int_\pi^0\log (iy) \log(1-e^{-2iy})dy=0$$ I'll call the sum of the first two integrals $I_1$ and the last one $I_2$. For $I_1$ we have: $$I_1=\int_0^\infty \log\bigg(\frac{x}{\sqrt{x^2+\pi^2}}\bigg)\log(1-e^{-2x})dx-i\int_0^\infty \arg(x+i\pi)\log(1-e^{-2x})dx$$ Of which, we can see, the real part is very interesting. Now, for $I_2$ we have $$I_2=-i\int_0^\pi\log(iy)\log(1-e^{-2iy})dy=-i\int_0^\pi \bigg[\log y+i\frac{\pi}{2}\bigg]\bigg[\log(2-2\cos 2y)+i \arg (1-e^{-2iy})\bigg]dy$$ Of which we are interested in the real part, which is: $$\Re I_2=\frac{\pi}{2}\int_0^\pi \log(2-2\cos 2y)dy + \int_0^\pi \log y \arg (1-e^{-2iy})dy=J_1+J_2$$ Let's evaluate the first one. $$J_1=\frac{\pi}{2}\int_0^\pi \log(2-2\cos 2y)dy=\frac{\pi}{2}\int_0^\pi \log(4\sin^2 y)dy=0$$ The last line follows from the well known integral of $\log(\sin x)$. Let's evaluate $J_2$. $$1-e^{-2iy}=1-\cos (2y) + i \sin (2y)= 2 \sin^2 y + 2i \sin y \cos y=2 \sin y(\sin y +i \cos y)=2i \sin y e^{-iy}$$ We can now see that, up to unimportant costants, the multiplication by $2i\sin y$ is a rotation by $\pi/2$, so in terms of argument it's equivalent to a multiplication by $e^{i\pi/2}$. So: $$\arg(1-e^{-2iy})=\arg (e^{i(\pi/2-y)})=\frac{\pi}{2}-y$$ So $J_2$ becomes: $$J_2=\int_0^\pi \log y\frac{\pi}{2}dy-\int_0^{\pi}y\log y dy=-\frac{\pi^2}{4}$$ The last line follows from elementary integration techniques. Putting everything together we have: $$\Re I_1=-\Re I_2$$ $$\int_0^\infty \log\bigg(\frac{x}{\sqrt{x^2+\pi^2}}\bigg)\log(1-e^{-2x})dx = \frac{\pi^2}{4} $$ So, multiplying by $2$ and remembering the original $1/2$ we have: $$I=\frac{\pi^2}{4}$$ As desired.

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  • $\begingroup$ Please tell me if something isn't clear in my post. $\endgroup$ – Gennaro Marco Devincenzis Oct 30 '14 at 16:23
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First, let $-\log(\sin(\theta))=y$ that yields $$\underbrace{\int_0^{\infty} \log(y) \log(1-e^{-2 y}) \ dy}_{\displaystyle \pi^2 \log(A)-\frac{1}{12}\pi^2 \log(\pi)}-\frac{1}{2}\underbrace{\int_0^{\infty} \log(\pi^2+y^2) \log(1-e^{-2 y}) \ dy}_{\displaystyle\sum_{k=1}^{\infty} \frac{\operatorname{Ci}(2k \pi)}{k^2} -\sum_{k=1}^{\infty}\frac{\log(\pi)}{k^2}}$$

where the series result is got by combining the series of $\log(1-e^{-2 y})$ and the exponential integral.
Making use of the $1.22 a$ from http://arxiv.org/pdf/1008.0040.pdf, we conclude that

$$\begin{equation}\large{\int_0^{\Large\frac{\pi}{2}}}\ln\left(\frac{\ln^2\sin\theta}{\pi^2+\ln^2\sin\theta}\right)\,\frac{\ln\cos\theta}{\tan\theta}\,d\theta=\frac{\pi^2}{4}\end{equation}$$

Q.E.D. (note I only used well-known old results - isn't it too easy for a contest?)

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    $\begingroup$ I apologize for not following the rule ... $\endgroup$ – user 1357113 Oct 26 '14 at 10:25
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    $\begingroup$ google.com.au/… $\endgroup$ – user142198 Oct 26 '14 at 10:31
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    $\begingroup$ @Committingtoachallenge Alternatively, you can ask Wolfram Alpha. Many of us probably already have that window open anyway. :) $\endgroup$ – David H Oct 26 '14 at 11:03
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    $\begingroup$ @Chris'ssis Well, brilliant.org is generally aimed at high school level students from what I have seen. I don't think it's too easy in that context. $\endgroup$ – Chantry Cargill Oct 26 '14 at 11:06
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    $\begingroup$ While the two resultant integrals aren't that difficult, it would certainly be nice if you could add in slightly more detail if you wish to. +1 $\endgroup$ – M.N.C.E. Oct 26 '14 at 13:29

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