# 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?

• 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. May 8, 2011 at 13:00
• Yes, the dilogarithm seems to be required here... May 8, 2011 at 13:03
• @Kolya: Do you actually want to compute $\int_a^b {\log (\sin (x))\,{\rm d}x}$ for certain $a$ and $b$? May 8, 2011 at 13:32
• For $a=0$ and $b=\pi/2$ or $b=\pi$, for example...
– Did
May 8, 2011 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.
– JSCB
Jul 12, 2012 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.

• 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. May 8, 2011 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. May 8, 2011 at 17:45
• Yes, and in Abramowitz and Stegun, too. May 8, 2011 at 17:46
• I was wondering just last night whether $$\int_{0}^{\pi/2}\ln^{k}(\sin{x})\;{dx}$$ where $k\in\mathbb{N}$, can be calculated! May 8, 2011 at 18:52
• @Lyrebird: The integrals you mentioned in your comment can be calculated by considering the parametric integral $\int_{0}^{\pi/2}\sin^t(x)\,\mathrm dx$, $t\ge 0$. Differentiating this integral $k$-times w.r.t. $t$ and letting $t\to 0^+$ generates integrals involving powers of logarithms. On the other hand, the parametric sine-integral with $t$ can be evaluated as a specific value of the Beta function. Expanding it around $t=0$ and considering its coefficients gives the required evaluation. Further generalizations are possible. Also, there is a connection to multiple-zeta-like values. Sep 23, 2021 at 17:39

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

• 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, 2014 at 14:43
• @hunminpark, How did you derive that identitiy in the beginning of this answer? Dec 15, 2014 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}

• You obtain the equation $2I=-\frac\pi2\cdot\ln 2+I$ and deduce that $I=-\frac\pi2\cdot\ln 2$. However, this equation is also satisfied for $I=\pm\infty$. Of course, $I=+\infty$ is absurd. But what about $I=-\infty$. I think that it is necessary to discuss the convergence of the original improper integral. Anyway, nice manipulations. (+1) Sep 22, 2021 at 20:24

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*}

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

• 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 $\alpha$... Jul 12, 2012 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, 2012 at 8:13

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 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: $$\int_0^x\log(\sin t) dt=\int_0^x\left[\log(2\sin t)-\log2\right] dt =-\frac12\Im[\operatorname{Li}_2(e^{2ix})] -x\log2.$$

$$\int\ln(\sin x)dx=\int\ln\left(\frac{(1-e^{-2ix})e^{ix}}{2i}\right)dx=\int\left(ix-\ln2-\frac{\pi i}{2}\right)dx+\int\ln(1-e^{-2ix})dx$$ The first integral on the RHS is easy to handle $$\int\left(ix-\ln2-\frac{\pi i}{2}\right)dx=\frac{ix^2}{2}-\left(\ln2+\frac{\pi i}{2}\right)x+C$$ We will use substitution to handle the second integral $$u=e^{-2ix}$$ $$dx=-\frac{du}{2iu}$$ $$\int\ln(1-e^{-2ix})dx=\frac{-1}{2i}\int\frac{\ln(1-u)}{u}du=\frac{\operatorname{Li}_2(u)}{2i}+C=\frac{\operatorname{Li}_2(e^{-2ix})}{2i}+C$$ We get $$\int\ln(\sin x)dx=\frac{ix^2}{2}-\left(\ln2+\frac{\pi i}{2}\right)x+\frac{\operatorname{Li}_2(e^{-2ix})}{2i}+C$$

• This does not look so real tough... Dec 26, 2022 at 16:38
• @BobDobbs what do you mean? i do use imaginary numbers i guess Jan 21, 2023 at 0:01

It seems that the integral can not be evaluated by elemantary functions, but one have to write the solution by using a special function which we have not seen in Calculus $$1$$. Instead of starting with the product expansion of $$\frac{\sin x}{x}$$, I saw the following Calculus $$1$$ trick: $$\ln\sin x=\ln\sin x+x\cot x-1+2\frac{1-x\cot x}{2}$$ Hence, $$\int\ln\sin x dx=x\ln\sin x-x+2\pi M\left(\frac{x}{\pi}\right)+c$$ where $$M(t)=\int_0^t\frac{1-\pi t\cot(\pi t)}{2}dt=\sum_{n=1}^{\infty}\frac{\zeta(2n)}{2n+1}t^{2n+1},\hspace{1cm} |t|<1.$$ Then, we have $$M(\frac{1}{2})=\frac14-\frac14\ln2$$. You may see WA output.

Hence, $$\int_0^{\pi/2}\ln\sin x dx=-\frac\pi 2+2\pi(\frac14-\frac14\ln2)=-\frac\pi 2\ln 2$$.

You may argue that for the computation of $$M(\frac{1}{2})$$, you will need $$\int_0^{\pi/2}\ln\sin x dx$$, so this is a chicken and egg situation. But, I think there are other methods to compute the values of $$M(x)$$ involving Gamma, Beta etc. functions.

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.

• Simply stating a closed form without a derivation seems mostly useless. Aug 18, 2014 at 23:40
• @Downvoter: What's the down vote for? Aug 18, 2014 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. Aug 18, 2014 at 23:43
• @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 $\mathrm{MickeyMouse} (x) = \int_0^x \log(\sin x) \, dx$... Apr 12, 2019 at 19:43