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?

  • 1
    $\begingroup$ 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. $\endgroup$ May 8 '11 at 13:00
  • $\begingroup$ Yes, the dilogarithm seems to be required here... $\endgroup$ May 8 '11 at 13:03
  • $\begingroup$ @Kolya: Do you actually want to compute $\int_a^b {\log (\sin (x))\,{\rm d}x}$ for certain $a$ and $b$? $\endgroup$
    – Shai Covo
    May 8 '11 at 13:32
  • 2
    $\begingroup$ For $a=0$ and $b=\pi/2$ or $b=\pi$, for example... $\endgroup$
    – Did
    May 8 '11 at 13:59
  • 1
    $\begingroup$ 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. $\endgroup$
    – JSCB
    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.

  • $\begingroup$ 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. $\endgroup$ May 8 '11 at 17:41
  • 1
    $\begingroup$ 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. $\endgroup$ May 8 '11 at 17:45
  • $\begingroup$ Yes, and in Abramowitz and Stegun, too. $\endgroup$ May 8 '11 at 17:46
  • $\begingroup$ (should have said he/she above. The ability to edit comments runs out too quickly.) $\endgroup$ May 8 '11 at 17:50
  • 2
    $\begingroup$ I was wondering just last night whether $$ \int_{0}^{\pi/2}\ln^{k}(\sin{x})\;{dx}$$ where $k\in\mathbb{N}$, can be calculated! $\endgroup$
    – Lyrebird
    May 8 '11 at 18:52

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 ... $\displaystyle K=\sum_{k\ge 1}\frac{(-1)^{k-1}}{(2k-1)^2}$)

  • $\begingroup$ 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). $\endgroup$
    – robjohn
    Mar 11 '14 at 14:43
  • 1
    $\begingroup$ @hunminpark, How did you derive that identitiy in the beginning of this answer? $\endgroup$
    – Amad27
    Dec 15 '14 at 7:57

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 $$ \int_0^{\pi/2}\log(\sin(x))\,dx=-\frac{\pi}{2}\log(2). $$


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.


$$ \begin{align} &I=\int_0^{\pi/2}\log(\sin x)dx=\int_0^{\pi/2}\log(\cos x)dx\\ \implies 2I&=\int_0^{\pi/2}\log(\sin x\cos x)dx=\int_0^{\pi/2}\log(\frac{1}{2}.2\sin x\cos x)dx\\ &=\int_0^{\pi/2}\log(1/2)dx+\int_0^{\pi/2}\log(\sin 2x)dx\\ &\text{Put }t=2x\implies dt=2dx\\ 2I&=\frac{\pi}{2}\log(\frac{1}{2})+\frac{1}{2}\int_0^{\pi}\log(\sin t)dt=\frac{-\pi}{2}\log 2+\frac{1}{2}\int_0^{\pi}\log(\sin x)dx\\ &=-\frac{\pi}{2}\log 2+\frac{1}{2}\int_0^{\pi/2}\log(\sin x)dx+\frac{1}{2}\int_0^{\pi/2}\log(\cos x)dx=-\frac{\pi}{2}\log 2+I\\ &\boxed{I=-\frac{\pi}{2}\log 2} \end{align} $$


(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 $\displaystyle \int_0^{\pi/2} \sin^{n}(x) \log(\sin(x)) dx$.

  • $\begingroup$ 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 $\alpha$... $\endgroup$ Jul 12 '12 at 8:11
  • $\begingroup$ @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. $\endgroup$
    – user17762
    Jul 12 '12 at 8:13

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*}


Probably this question indeed aims on a definite integral from $0$ to $\pi/2$ (or $\pi$). But it can be of interest to give a rather simple form of the antiderivative in the range $0<x<\pi$ as well.

Lemma: $$ \int_0^x\log(2\sin t) dt=-\frac12\sum_{k=1}^\infty\frac{\sin(2kx)}{k^2} =-\frac12\Im[\operatorname{Li}_2(e^{2ix})],\tag1 $$ Proof:

Observe that $\log(2\sin t)$ in the range $0< t<\pi$ is a real number. Therefore: $$\begin{align} \log(2\sin t)&=\log(e^{it}-e^{-it})-\log i\\ &=i\left(t-\frac\pi2\right)+\log(1-e^{-2it})\\ &=i\left(t-\frac\pi2\right)-\sum_{k=1}^\infty\frac{e^{-2ikt}}k\\ &=i\underbrace{\left(t-\frac\pi2+\sum_{k=1}^\infty\frac{\sin(2kt)}k\right)}_{=0}-\sum_{k=1}^\infty\frac{\cos(2kt)}k\\ &=-\sum_{k=1}^\infty\frac{\cos(2kt)}k.\tag2 \end{align} $$ Substituting $(2)$ into the left hand side of $(1)$ one obtains its right hand side. $\blacksquare$

Thus for the integral in question we have: $$ \int_0^x\log(\sin t) dt=\int_0^x\left[\log(2\sin t)-\log2\right] dt =-\frac12\Im[\operatorname{Li}_2(e^{2ix})] -x\log2. $$


For the indefinite integral, you 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 a polylogarithm.

  • 8
    $\begingroup$ Simply stating a closed form without a derivation seems mostly useless. $\endgroup$ Aug 18 '14 at 23:40
  • $\begingroup$ @Downvoter: What's the down vote for? $\endgroup$ Aug 18 '14 at 23:41
  • 3
    $\begingroup$ @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. $\endgroup$ Aug 18 '14 at 23:43
  • 2
    $\begingroup$ @MhenniBenghorbal Using polylogarithms is really not useful here... I think it is even debatable if this can be called a closed form solution. You can always come up with new special functions, name them, and say you have a closed form solution. In this case we could just define $\mathrm{MickeyMouse} (x) = \int_0^x \log(\sin x) \, dx$... $\endgroup$ Apr 12 '19 at 19:43

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.