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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
    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.2012-07-12

8 Answers 8

34

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.

  • 2
    I was wondering just last night whether $ \int_{0}^{\pi/2}\ln^{k}(\sin{x})\;{dx}$ where $k\in\mathbb{N}$, can be calculated!2011-05-08
14

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

  • 0
    @hunminpark, How did you derive that identitiy in the beginning of this answer?2014-12-15
9

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.

8

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). $

4

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

  • 0
    @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.2012-07-12
4

$ \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} $

3

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*}

-4

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.