Computing the integral of $\log(\sin x)$

How to compute the following integral? $$\int\log(\sin x)\,dx$$

Motivation: Since $\log(\sin x)'=\cot x$, the antiderivative $\int\log(\sin x)\,dx$ has the nice property $F''(x)=\cot x$. Can we find $F$ explicitly? Failing that, can we find the definite integral over one of intervals where $\log (\sin x)$ is defined?

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I'm pretty sure this is an integral that can't be expressed in terms of elementary functions (that is, the functions of 1st-year calculus). See, for example, reference.wolfram.com/legacy/v5/TheMathematicaBook/… about halfway down the page. –  Gerry Myerson May 8 '11 at 13:00
Yes, the dilogarithm seems to be required here... –  Guess who it is. May 8 '11 at 13:03
@Kolya: Do you actually want to compute $\int_a^b {\log (\sin (x))\,{\rm d}x}$ for certain $a$ and $b$? –  Shai Covo May 8 '11 at 13:32
For $a=0$ and $b=\pi/2$ or $b=\pi$, for example... –  Did May 8 '11 at 13:59
Although this integral may cannot be expressed in elementary function, but it may can be expressed in series form. For example, ∫sin(sin x)dx and ∫cos(cos x)dx can both be evaluated in series form. –  ᴊ ᴀ s ᴏ ɴ Jul 12 '12 at 8:21

You can calculate $$\int_0^\pi\log(\sin x)\,dx = -\pi\log2$$ and integrating up to $\pi/2$ would give half of this.

Note that integrating $\log(\sin x)$ from 0 to $\pi/2$ is the same as integrating $\log(\cos x)$ so that \begin{align} \int_0^{\pi/2}\log(\sin x)\,dx &= \frac12\int_0^{\pi/2}\log(\sin x\cos x)\,dx\\ &= \frac12\int_0^{\pi/2}\log(\sin 2x)\,dx - \frac{\pi}{4}\log 2. \end{align} After a change of variables, this can be rearranged to get the result.

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Actually, as the OP hasn't come back to say if it was the definite or indefinite integral that he was after, I'm not sure if this fully answers the question. –  George Lowther May 8 '11 at 17:41
Also, I'm not sure what the appropriate amount of detail is for a homework question. The value of the integral is no secret anyway, as Wolfram alpha knows it. –  George Lowther May 8 '11 at 17:45
Yes, and in Abramowitz and Stegun, too. –  Guess who it is. May 8 '11 at 17:46
(should have said he/she above. The ability to edit comments runs out too quickly.) –  George Lowther May 8 '11 at 17:50
I was wondering just last night whether $$\int_{0}^{\pi/2}\ln^{k}(\sin{x})\;{dx}$$ where $k\in\mathbb{N}$, can be calculated! –  Lyrebird May 8 '11 at 18:52

I think it worth mentioning the history of (essentially) this function, tracing back to work of Lobachevsky in the beginnings of non-Euclidean geometry. See the pdf here for Milnor's survey, the function is discussed from page 9 onward.

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Series expansion can be used for this integral too.
We use the following identity; $$\log(\sin x)=-\log 2-\sum_{k\geq 1}\frac{\cos(2kx)}{k} \phantom{a} (0<x<\pi)$$ This identity gives $$\int_{a}^{b} \log(\sin x)dx=-(b-a)\log 2-\sum_{k\ge 1}\frac{\sin(2kb)-\sin(2ka)}{2k^2}$$ ($a, b<\pi$)
For example, $$\int_{0}^{\pi/4}\log(\sin x)dx=-\frac{\pi}{4}\log 2-\sum_{k\ge 1}\frac{\sin(\pi k/2)}{2k^2}=-\frac{\pi}{4}\log 2-\frac{1}{2}K$$ $$\int_{0}^{\pi/2} \log(\sin x)dx=-\frac{\pi}{2}\log 2$$ $$\int_{0}^{\pi}\log(\sin x)dx=-\pi \log 2$$ ($K$; Catalan's constant ... $\displaystyle K=\sum_{k\ge 1}\frac{(-1)^{k-1}}{(2k-1)^2}$)

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I discovered the identity you used above as $\sin^2(x)=\dfrac{1-\cos(2x)}{2}=\dfrac{(1-e^{2ix})(1-e^{-2ix})}{4}$ while answering this question. I was lead here via a series of links. Nice answer (+1). –  robjohn Mar 11 '14 at 14:43
@hunminpark, How did you derive that identitiy in the beginning of this answer? –  Amad27 Dec 15 '14 at 7:57

(I am assuming that the OP is interested in the definite integral).

The following argument is not completely rigorous $\displaystyle \int_0^{\pi/2} \log(\sin(x)) dx = - \dfrac{\pi}2 \log 2$ but I think it can be made rigorous.

From integration by parts/ other techniques, we have that $$\int_0^{\pi/2} \sin^{2k}(x) dx = \frac{2k-1}{2k}\frac{2k-3}{2k-2} \cdots \frac{1}{2} \frac{\pi}{2} = \dfrac{(2k)!}{4^k (k!)^2} \dfrac{\pi}2 = \dfrac{\Gamma(2k+1)}{4^k \Gamma^2(k+1)} \dfrac{\pi}2$$

Hence, a possible analytic extension to $\displaystyle \int_0^{\pi/2} \sin^{2z}(x) dx$ is $\dfrac{\Gamma(2z+1)}{4^z \Gamma^2(z+1)} \dfrac{\pi}2$.

Now differentiate both sides with respect to $z$, and set $z=0$, to get $$2 \int_0^{\pi/2} \log(\sin(x)) = -\dfrac{\pi}2 \log(4)$$ Hence, we get that $$\int_0^{\pi/2} \log(\sin(x)) dx = -\dfrac{\pi}2 \log(2)$$ This also provides you a way to evaluate $\displaystyle \int_0^{\pi/2} \sin^{n}(x) \log(\sin(x)) dx$.

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The differentiation under the integral sign is fine, I think, so it seems to me that the only gap is to justify the expression for $\int_0^{\pi/2} \sin^{2\alpha}(x)\mathrm dx$ for noninteger $\alpha$... –  Guess who it is. Jul 12 '12 at 8:11
@J.M. Actually thinking about it since the domain is only from $0$ to $\pi/2$, $\sin^{2 \alpha}(x)$ is well defined even for non-integer $\alpha$. So I think this does it. Hence, the analytic extension is the analytic extension. –  user17762 Jul 12 '12 at 8:13

An excellent discussion of this topic can be found in the book The Gamma Function by James Bonnar. Consider just two of the provably equivalent definitions of the Beta function: $$\begin{eqnarray} B(x,y)&=& 2\int_0^{\pi/2}\sin(t)^{2x-1}\cos(t)^{2y-1}\,dt\\ &=& \frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}. \end{eqnarray}$$

Directly from this definition we have

$$B(n+\frac{1}{2},\frac{1}{2}): \int_0^{\pi/2}\sin^{2n}(x)\,dx=\frac{\sqrt{\pi} \cdot\Gamma(n+1/2)}{2(n!)}$$ $$B(n+1,\frac{1}{2}): \int_0^{\pi/2}\sin^{2n+1}(x)\,dx=\frac{\sqrt{\pi} \cdot n!}{2 \Gamma(n+3/2)}$$ Hence the quotient of these two integrals is $$\begin{eqnarray} \frac{ \int_0^{\pi/2}\sin^{2n}(x)\,dx}{\int_0^{\pi/2}\sin^{2n+1}(x)\,dx}&=& \frac{\Gamma(n+1/2)}{n!}\frac{\Gamma(n+3/2)}{n!}\\ &=& \frac{2n+1}{2n}\frac{2n-1}{2n}\frac{2n-1}{2n-2}\cdots\frac{3}{4}\frac{3}{2}\frac{1}{2}\frac{\pi}{2} \end{eqnarray}$$ where the quantitiy $\pi/2$ results from the fact that $$\frac{\int_0^{\pi/2}\sin^{2\cdot 0}(x)\,dx}{\int_0^{\pi/2}\sin^{2\cdot 0+1}x\,dx}=\frac{\pi/2}{1}=\frac{\pi}{2}.$$ So we have that $$\int_0^{\pi/2}\sin^{2n}(x)\,dx=\frac{2n-1}{2n}\frac{2n-3}{2n-2}\cdots\frac{1}{2}\frac{\pi}{2}=\frac{(2n)!}{4^n (n!)^2}\frac{\pi}{2}.$$ Hence an analytic continuation of $\int_0^{\pi/2}\sin^{2n}(x)\,dx$ is $$\int_0^{\pi/2}\sin^{2z}(x)\,dx=\frac{\pi}{2}\frac{\Gamma(2z+1)}{4^z \Gamma^2(z+1)}=\frac{\pi}{2}\Gamma(2z+1)4^{-z}\Gamma^{-2}(z+1).$$ Now differentiate both sides with respect to $z$ which yields

$$\begin{eqnarray} 2\int_0^{\pi/2}\sin^{2z}(x)\log(\sin(x))\,dx =\frac{\pi}{2} \{2\Gamma'(2z+1)4^{-z}\Gamma^{-2}(z+1)\\ +2\Gamma(2z+1)4^{-z}\Gamma^{-3}(z+1)\Gamma'(z+1)\\ -\log(4)\Gamma(2z+1)4^{-z}\Gamma^{-2}(z+1)\}. \end{eqnarray}$$

Finally set $z=0$ and note that $\Gamma'(1)=-\gamma$ to complete the integration: $$\begin{eqnarray} 2\int_0^{\pi/2}\log(\sin(x))\,dx&=&\frac{\pi}{2}(-2\gamma+2\gamma-\log(4))\\ &=& -\frac{\pi}{2}\log(4)=-\pi\log(2). \end{eqnarray}$$ We conclude that $$\int_0^{\pi/2}\log(\sin(x))\,dx=-\frac{\pi}{2}\log(2).$$

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Nice solution..... –  juantheron May 31 '14 at 4:01

There was a duplicate posted a while ago. Since I think my answer might be of some interest, here it goes:

By substituting $\sin{x}=t$, we can write it as: \begin{align*} \int_{0}^{\pi/2} \, \log\sin{x}\, dx &= \int_{0}^{1} \, \frac{\log{t}}{\sqrt{1-t^2}}\, dt \tag{1} \end{align*}

Now, consider:

\begin{align*} I(a) &= \int_{0}^{1} \, \frac{t^a}{(1-t^2)^{1/2}}\, dt \\ &= \mathrm{B}\left(\frac{a+1}{2},\; \frac{1}{2}\right) \\ \frac{\partial }{\partial a}I(a) &= \frac{1}{4}\left(\psi\left(\frac{a+1}{2}\right)-\psi\left(\frac{a+2}{2}\right)\right)\mathrm{B}\left(\frac{a+1}{2},\; \frac{1}{2}\right) \\ \implies I'(0) &= \frac{1}{4}\left(\psi\left(\frac{1}{2}\right)-\psi\left(1\right)\right)\mathrm{B}\left(\frac{1}{2},\; \frac{1}{2}\right) \tag{2} \end{align*} Putting the values of digamma and beta functions. \begin{align*} \psi\left(\frac{1}{2}\right) &= -2\log{2}-\gamma \\ \psi\left(1\right) &= -\gamma \\ \mathrm{B}\left(\frac{1}{2}, \frac{1}{2}\right) &= \pi \end{align*}

Hence, from $(1)$ and $(2)$, \begin{align*} \boxed{\displaystyle \int_{0}^{\pi/2} \, \log\sin{x}\, dx = -\frac{\pi}{2}\log{2}} \end{align*}

Using a CAS, we can derive for higher powers of $\ln\sin{x}$, e.g. \begin{align*} \int_{0}^{\pi/2} \, \left(\log\sin{x}\right)^2\, dx &= \frac{1}{24} \, \pi^{3} + \frac{1}{2} \, \pi \log\left(2\right)^{2} \\ \int_{0}^{\pi/2} \, \left(\log\sin{x}\right)^3\, dx &= -\frac{1}{8} \, \pi^{3} \log\left(2\right) - \frac{1}{2} \, \pi \log\left(2\right)^{3} - \frac{3}{4} \, \pi \zeta(3)\\ \int_{0}^{\pi/2} \, \left(\log\sin{x}\right)^4\, dx &= \frac{19}{480} \, \pi^{5} + \frac{1}{4} \, \pi^{3} \log\left(2\right)^{2} + \frac{1}{2} \, \pi \log\left(2\right)^{4} + 3 \, \pi \log\left(2\right) \zeta(3) \end{align*}

We can also observe another interesting thing, for small values of $n$
\begin{align*} \displaystyle \int_{0}^{\pi/2} \, \left(\log\sin{x}\right)^n\, dx \approx \displaystyle (-1)^n\, n! \end{align*}

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The answer is correct and it is the only answer that matches the OP question since he is asking for the evaluation of indefinite integral

You can have this closed form

$$\frac{i{x}^{2}}{2}+x\ln \left( \cos \left( x \right) \right) -x\ln \left( 1+{{\rm e}^{2\,ix}} \right) +\frac{i}{2} Li_2 ( -{ {\rm e}^{2\,ix}} ),$$

where $Li_2$ is polyloagarith.

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Simply stating a closed form without a derivation seems mostly useless. –  Carl Mummert Aug 18 '14 at 23:40
@Downvoter: What's the down vote for? –  Mhenni Benghorbal Aug 18 '14 at 23:41
@CarlMummert: It tells people there exists a closed form and whoever is interested in proving it can put some effort to find it. Giving detailed answers all the time is not useful. –  Mhenni Benghorbal Aug 18 '14 at 23:43