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

Cameron Williams
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Anastasiya-Romanova 秀
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    I have purged the comments to this question, which all deal with meta-issues best handled on the related [meta-thread](http://meta.math.stackexchange.com/questions/17136/using-math-se-as-a-contest-site). – user642796 Oct 25 '14 at 09:18
  • I edited out the contest bits because they're intrusive and the contest is long since expired at this point. It can still be found in the history of the question for anyone who is interested. – Cameron Williams Nov 25 '21 at 17:11

4 Answers4


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.

Sangchul Lee
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    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 – Anastasiya-Romanova 秀 Oct 26 '14 at 12:18
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    @Anastasiya-Romanova, Thank you! And I just found that this approach is quite similar to that of the [solution of Eric Naslund](http://math.stackexchange.com/questions/523027/a-math-contest-problem-int-01-ln-left1-frac-ln2x4-pi2-right-frac/543760#543760) for a similar problem. – Sangchul Lee Oct 26 '14 at 12:26
  • 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](https://brilliant.org/discussions/thread/integral-contest/) and [Quora](http://qr.ae/DKomq) in case you wanna answer it there – Anastasiya-Romanova 秀 Oct 26 '14 at 12:39

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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    Brilliant answer as usual & more important it is easy to understand for a high school student like me, +1 ≧◠‿◠≦✌ – Anastasiya-Romanova 秀 Oct 27 '14 at 15:23
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    @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! ^_^; – David H Oct 30 '14 at 17:47

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.


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


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

user 1591719
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