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I am trying to compute this integral.

$$\int_0^{\frac{\pi}{2}} \frac{\ln(\sin(x))}{\cot(x)}\ \text{d}x$$

Any thoughts will help. Thanks.

Jendrik Stelzner
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mnm
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    Firstly one has to find out whether it converges. In 0, the function is bounded since it is infinitesimal. What about $\frac\pi2$? It is asymptotic to $\frac{\sin x-1}{\cos x}$ there. By L'Hôpital's rule this goes to $\lim_{x\to\frac\pi2}\frac{\cos x}{-\sin x}=0$, so OK, the integral converges. – MickG Jan 05 '16 at 21:04
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    Let $u=\sin x$, and you get the integral $$\int_0^1\frac{\ln u}{2(1-u)}-\frac{\ln u}{2(1+u)},du.$$ Just turn the corresponding geometric series around and integrate. You will get a rational constant times $\pi^2$. – mickep Jan 05 '16 at 21:11
  • @mickep nice, that was also my first thaught – tired Jan 05 '16 at 22:19

4 Answers4

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Note that

$$\int_{0}^{\frac{\pi}{2}}{\frac{\ln(\sin{x})}{\cot{x}}}\,dx = \int_{0}^{\frac{\pi}{2}}{\ln(\sin{x})\tan{x}}\,dx = \int_{0}^{\frac{\pi}{2}}\frac{\ln(\sin{x})\sin{x}}{\cos{x}}\,dx $$ $$= \int_{0}^{\frac{\pi}{2}}\frac{\ln\left(\sqrt{1-\cos^2{x}}\right)\sin{x}}{\cos{x}}\,dx$$

Now, set $u=\cos{x}\implies du=-\sin{x}\,dx$. Also, note that $\ln\left(\sqrt{1-u^2}\right)=\frac{1}{2}\ln\left(1-u^2\right)$ Also, don’t forget to change the limits. Now, we have

$$\int_{0}^{\frac{\pi}{2}}\frac{\ln\left(\sqrt{1-\cos^2{x}}\right)\sin{x}}{\cos{x}}\,dx=\frac{1}{2}\int_{0}^{1}\frac{\ln\left(1-u^2\right)}{u}\,du$$

There are two paths you can take from here. One I have shown below, and the other is shown by Dr. MV in the comments below.

Now, we need to find what $\ln\left(1-u^2\right)$ is. Recall the Maclaurin Series for a natural logarithm. Therefore,

$$\ln\left(1-u^2\right)=-\sum_{n=1}^{\infty}\frac{u^{2n}}{n}$$

$$\therefore \frac{1}{2}\int_{0}^{1}\frac{\ln\left(1-u^2\right)}{u}\,du=-\frac{1}{2}\int_{0}^{1}\sum_{n=1}^{\infty}\frac{u^{2n}}{nu}\,du=-\frac{1}{2}\sum_{n=1}^{\infty}\int_{0}^{1}\frac{u^{2n-1}}{n}\,du$$ $$=-\frac{1}{4}\sum_{n=1}^{\infty}\left.\frac{u^{2n}}{n^2}\right|_{u=0}^{u=1}=-\frac{1}{4}\sum_{n=1}^{\infty}\frac{1}{n^2}$$ Since $\sum_{n=1}^{\infty}\frac{1}{n^2}=\frac{\pi^2}{6}$, $$\int_{0}^{\frac{\pi}{2}}{\frac{\ln(\sin{x})}{\cot{x}}}\,dx=-\frac{\pi^2}{24}$$

  • One thing: how do you justify swapping integral and series? – MickG Jan 05 '16 at 21:12
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    $$-\int_0^1 \frac{\log(1-u^2)}{u},du=\text{Li}_2(1)+\text{Li}_2(-1)=\frac12 \text{Li}_2(1)=\frac{\pi^2}{12}$$ – Mark Viola Jan 05 '16 at 21:13
  • @Dr.MV If you want to go the polylogarithmic way, it seems that $$-(\ln \sin x)(\ln(1+\sin x))/2+\text{Li}_2(1-\sin x)/2-\text{Li}_2(-\sin x)/2$$ is a primitive to the original integrand. – mickep Jan 05 '16 at 21:16
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    @MickG Monotone Convergence Theorem : all the terms of the series are non-negative because $0\leq u\leq1$. – Nicolas Jan 05 '16 at 21:16
  • @MickG What Nicolas said is the justification, but this may be of help too http://math.stackexchange.com/questions/83721/when-can-a-sum-and-integral-be-interchanged –  Jan 05 '16 at 21:21
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    @mickep I don't see a compelling reason to compute the anti-derivative in this case. I was pointing out that there is no need to proceed with the series representation after the substitution when the integral is already in the form that leads to a very straightforward result. – Mark Viola Jan 05 '16 at 21:21
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Note that \begin{eqnarray} &&\int_0^{\frac{\pi}{2}} \frac{\ln(\sin(x))}{\cot(x)}\ \text{d}x\\ &=&\int_0^{\frac{\pi}{2}}(\cos x)^{-1}\sin x\ln(\sin x)\ \text{d}x=\lim_{a\to0,b\to2}\frac{d}{db}\int_0^{\frac{\pi}{2}}(\cos x)^{a-1}(\sin x)^{b-1}\ \text{d}x\\ &=&\lim_{a\to0,b\to2}\frac{d}{db}\frac{1}{2}B\left(\frac{a}{2},\frac{b}{2}\right)\\ &=&\lim_{a\to0,b\to2}\frac{1}{4}B\left(\frac{a}{2},\frac{b}{2}\right)\left(\psi\left(\frac{b}{2}\right)-\psi\left(\frac{a+b}{2}\right)\right)\\ &=&\lim_{a\to0}\frac{1}{4}B\left(\frac{a}{2},1\right)\left(\psi\left(\frac{b}{2}\right)-\psi\left(\frac{a+2}{2}\right)\right)\\ &=&\lim_{a\to0}\frac{1}{4}B\left(\frac{a}{2},1\right)\left(\psi(1)-\psi\left(\frac{a}{2}+1\right)\right)\\ &=&-\lim_{a\to0}\frac{1}{4}B\left(\frac{a}{2},1\right)\left(\gamma+\psi\left(\frac{a}{2}+1\right)\right)\\ &=&-\lim_{a\to0}\frac{1}{4}\frac{\Gamma(\frac{a}{2})\Gamma(1)}{\Gamma(\frac{a+2}{2})}\left(\gamma+\psi\left(\frac{a}{2}+1\right)\right)\\ &=&-\lim_{a\to0}\frac{1}{4}\left(\frac{2}{a}+O(a^2)\right)\left(\frac{\pi^2}{12}a+O(a^2)\right)\\ &=&-\frac{\pi^2}{24}. \end{eqnarray} Here we use the following fact $$ \Gamma(a)\approx\frac{1}{a}, \gamma+\psi(\frac{a}{2}+1)\approx\frac{\pi^2}{12}a.$$

MickG
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xpaul
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A partial answer: Let $I$ be the given integral. You can integrate by parts, letting $$\begin{matrix}u=\dfrac{\ln(\sin x)}{\cot x}&&&\mathrm{d}v=\mathrm{d}x\\[1ex] \mathrm{d}u=\left(1+\sec^2x\ln(\sin x)\right)\,\mathrm{d}x&&&v=x\end{matrix}$$ which yields $$\begin{align*} I&=\frac{x\ln(\sin x)}{\cot x}\Bigg|_{x=0}^{x=\pi/2}-\int_0^{\pi/2} x\left(1+\sec^2x\ln(\sin x)\right)\,\mathrm{d}x\\[1ex] &=0-\left(\int_0^{\pi/2}x\,\mathrm{d}x+\int_0^{\pi/2} x\sec^2x\ln(\sin x)\,\mathrm{d}x\right)\\[1ex] &=-\frac{\pi^2}{8}-\int_0^{\pi/2} x\sec^2x\ln(\sin x)\,\mathrm{d}x \end{align*}$$ It turns out that the remaining integral is twice $I$, which means $3I=-\dfrac{\pi^2}{8}$, or $I=-\dfrac{\pi^2}{24}$. I'm still seeing what can be done to establish this last part, though.

user170231
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Hint: $\cot(x) = (\ln(\sin(x)))'$

adjan
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  • Maybe I'm missing something, but would that not help if we had the reciprocal of that ratio, concluding the antiderivative was $\log(\log(\sin x))$? Besides, seeing Wolfram's antiderivative and its bunch of $Li_2$s, I find it hard to believe that such a simple hint would solve the problem… – MickG Jan 05 '16 at 20:59
  • I saw that too, but I don't know how it helps. Can you add some more details to your hint – mnm Jan 05 '16 at 21:00
  • It would be $\frac{f'(x)}{f(x)}$ where $f(x)=\cot x$ which is helpful –  Feb 20 '16 at 03:18