Evaluate: $ \displaystyle \int_0^{\pi} \ln \left( \sin \theta \right) d\theta$ using Gauss Mean Value theorem.

Given hint: consider $f(z) = \ln ( 1 +z)$.

EDIT:: I know how to evaluate it, but I am looking if I can evaluate it using Gauss MVT.

ADDED:: Here is what I have got so far!!

$$\ln 2 = \frac{1}{2 \pi } \int_0^{2\pi } \log(2+e^{i \theta}) d\theta = \frac{1}{2 \pi } \int_0^{2\pi } \log(2+e^{-i \theta}) d\theta$$

Hence, $ \displaystyle 2 \ln 2 = \frac{1}{2 \pi } \int_{0}^{2 \pi} \log(5 + 4 \cos \theta )d \theta = \frac{1}{\pi} \int_0^{\pi} \log(1 + 8 \cos^2 \theta) d \theta$, now to problem is how to reduce it to the above form?

  • $\begingroup$ But $\ln$ isn't an analytic function on any domain containing the circle around $1$ with radius $1$. Maybe this is a problem. $\endgroup$ – Cocopuffs Apr 8 '13 at 11:56
  • 1
    $\begingroup$ Similar question here: math.stackexchange.com/questions/37829/… $\endgroup$ – Cocopuffs Apr 8 '13 at 12:14
  • $\begingroup$ @Cocopuffs i am aware of this question. changing the angle to half angle. I am supposed to use Gauss MVT here. $\endgroup$ – Mula Ko Saag Apr 8 '13 at 12:15
  • 1
    $\begingroup$ Do you know the rule $\log(z) = \log|z| + i \operatorname{Arg} z$? This and the fact $|1 - e^{i \theta}| = |2 \sin \theta|$ may help. Also, are you from nepal? $\endgroup$ – muzzlator Apr 8 '13 at 13:24
  • $\begingroup$ @muzzlator yes that I am. $\endgroup$ – Mula Ko Saag Apr 8 '13 at 14:48

Here is a solution I wrote for a complex analysis assignment several years ago, I hope it helps. Basically, we are using the mean value theorem you mention above on a slightly different function, and then separating things to obtain the desired integral. We have to be careful because we can't exactly integrate $\log(1-u)$ on the circle of radius $1$.

Consider $$ \int_{C_{1-\epsilon}}\frac{\log(1-u)}{u}du $$ where $C_{1-\epsilon}$ is the circle of radius $1-\epsilon$. Then since $\frac{\log(1-u)}{u}$ is an analytic function in $D_{1-\epsilon}$ (It has a removable singularity at $u=0$ by the removable singularity theorem mentioned last assignment), we see that this contour integral will be zero for every $\epsilon>0$. But then notice $$ \int_{C_{1-\epsilon}}\frac{\log(1-u)}{u}du=2i\int_{0}^{\pi}\log(1-(1-\epsilon)e^{i2z})dz $$ so that $$ \int_{0}^{\pi}\log(1-(1-\epsilon)e^{i2z})dz=0 $$ for every $\epsilon>0$. Since $$ |\int_{0}^{\pi}\log(1-e^{i2z})dz|\leq\int_{0}^{\pi}|\log z|dz+\int_{0}^{\pi}|\log(\pi-z)|dz+\int_{0}^{\pi}|\log\left(\frac{1-e^{i2z}}{z(z-\pi)}\right)|dz $$ As $\frac{1-e^{i2z}}{z(z-\pi)}$ has no zeros on $[0,\pi]$ we see that it must be bounded below by some constant $c$. Then as it also has nontrivial imaginary part on $(0,\pi)$ we see that $\int_{0}^{\pi}|\log\left(\frac{1-e^{i2z}}{z(z-\pi)}\right)|dz<\infty$. Then since $\int_{0}^{1}\log xdx=x\log x-x\biggr|_{x=0}^{x=1}=-1<\infty$ it follows that $\int_{0}^{\pi}|\log z|dz<\infty$ and $\int_{0}^{\pi}|\log(\pi-z)|dz<\infty$ so that $|\int_{0}^{\pi}\log(1-e^{i2z})dz|<\infty$. Recall $\log$ is uniformly continuous on any compact set not containing the origin, so we can bound the middle of all of these integrals by the same constant. Since around $0$ and around $\pi$ the norm of $\log(1-e^{i2z})$ goes to infinity, we can choose small enough neighborhoods so that the norm of $\log(1-(1-\epsilon)e^{i2z})dz$ is bounded above by $|\log(1-e^{i2z})|$ in these neighborhoods for every $\epsilon>0$. Then applying the dominated convergence theorem tells us that $$ \lim_{\epsilon\rightarrow0}\int_{0}^{\pi}\log(1-(1-\epsilon)e^{i2z})dz=\int_{0}^{\pi}\log(1-e^{i2z})dz=0. $$ Now we have the identity $$ 1-e^{-2iz}=-2ie^{iz}\sin z $$ so that $$ 0=\int_{0}^{\pi}\log(\sin z))dz+\int_{0}^{\pi}\log(e^{iz})dz+\int_{0}^{\pi}\log(-2i)dz. $$ By choosing the principal branch of the logarithm we then have $$ \int_{0}^{\pi}\log(\sin z))dz=-\left(\int_{0}^{\pi}izdz+\int_{0}^{\pi}-\frac{\pi i}{2}dz+\int_{0}^{\pi}\log(2)dz\right) $$ $$ =-\left(\frac{i\pi^{2}}{2}+-\frac{\pi^{2}i}{2}dz+\pi\log(2)dz\right)=-\pi\log2. $$ By substituting $z=\pi x$ we see that $\int_{0}^{\pi}\log(\sin z))dz=\pi\int_{0}^{1}\log(\sin\pi x))dx$ so that we are able to conclude $$ \int_{0}^{1}\log(\sin\pi x))dx=-\log2 $$ as desired.

  • $\begingroup$ all right ... thanks!! $\endgroup$ – Mula Ko Saag May 8 '13 at 20:42

I got this as the first part of this answer:

Start with $$ \begin{align} \int_0^{\pi/2}\log(\sin(x))\,\mathrm{d}x &=\frac12\int_0^\pi\log(\sin(x))\,\mathrm{d}x\\ &=\int_0^{\pi/2}\log(\sin(2x))\,\mathrm{d}x\\ &=\int_0^{\pi/2}\Big(\log(2)+\log(\sin(x))+\log(\cos(x))\Big)\,\mathrm{d}x\\ &=\frac\pi2\log(2)+2\int_0^{\pi/2}\log(\sin(x))\,\mathrm{d}x\tag{1} \end{align} $$ Therefore, $$ \int_0^{\pi/2}\log(\sin(x))\,\mathrm{d}x=-\frac\pi2\log(2)\tag{2} $$

Thus, $$ \int_0^\pi\log(\sin(x))\,\mathrm{d}x=-\pi\log(2) $$

Using Gauss Mean Value


$\hspace{4.5cm}$enter image description here $$ \begin{align} \int_0^\pi\log(\sin(x))\,\mathrm{d}x &=\int_0^\pi\log\left(\color{#C00000}{\frac12}\sqrt{2-2\cos(x)}\right)\,\mathrm{d}x\\ &=\pi\color{#00A000}{\frac1{2\pi}\int_0^{2\pi}\log\left(\sqrt{2-2\cos(x)}\right)\,\mathrm{d}x}\color{#C00000}{-\pi\log(2)}\\[6pt] &=\pi\color{#00A000}{\log(1)}-\pi\log(2)\\[12pt] &=-\pi\log(2) \end{align} $$

  • 1
    $\begingroup$ (+1.5) the picture itself should worth an extra 1/2 upvote ;-p $\endgroup$ – achille hui May 2 '13 at 6:58
  • $\begingroup$ @robjohn I'm not understanding the first line. How does it follow that sin = 1/2 sqrt( 2 - 2 cos)? $\endgroup$ – Mark Dec 29 '14 at 19:45
  • 2
    $\begingroup$ @Mark: We use $\cos(2x)=1-2\sin^2(x)\implies\sin(x)=\sqrt{\frac{1-\cos(2x)}2}$. Then, by symmetry and this equation, $$\begin{align}\int_0^\pi\log(\sin(x))\,\mathrm{d}x &=2\int_0^{\pi/2}\log(\sin(x))\,\mathrm{d}x\\ &=2\int_0^{\pi/2}\log\left(\sqrt{\frac{1-\cos(2x)}2}\right)\,\mathrm{d}x\\ &=\int_0^\pi\log\left(\sqrt{\frac{1-\cos(x)}2}\right)\,\mathrm{d}x\end{align}$$ The last step is achieved by substituting $x\mapsto x/2$. $\endgroup$ – robjohn Dec 30 '14 at 1:06

This is just for fun.

It is well known that $\Pi_{1\le k<n}\sin \frac{k\pi}{n}=\frac{n}{2^{n-1}}$. So $\int_0^\pi \ln \sin x dx =\lim_{n\to \infty}\frac{\pi}{n}\ln \Pi_{1\le k<n} \sin \frac{k\pi}{n}=-\pi\ln 2 $.

  • $\begingroup$ very interesting!! i knew the identity $\endgroup$ – Mula Ko Saag May 8 '13 at 20:38

$$I=\displaystyle \int_0^{\pi} \ln \left( \sin \theta \right)\cdot d\theta$$ $$I=2\times \displaystyle \int_0^{\pi/2} \ln \left( \sin \theta \right) \cdot d\theta$$ $$I=2\times \displaystyle \int_0^{\pi/2} \ln \left( \cos \theta \right) \cdot d\theta$$ Adding both. $$I=\displaystyle \int_0^{\pi/2} \ln \left( \sin \theta \times \cos \theta\right) \cdot d\theta$$ $$I= \displaystyle \int_0^{\pi/2} \ln \left(2 \sin \theta\times \cos \theta \right) -\ln2 \cdot d\theta$$ $$I=\int_0^{\pi/2}\ln(\sin{2\theta})-\ln2 \cdot d\theta$$ $$\int_0^{\pi/2}\ln(\sin{2\theta})\cdot d\theta=I/2$$ So, $$I=-2\int_0^{\pi/2}\ln2\cdot d\theta$$ $$I=-{\pi\ln2}$$

  • $\begingroup$ @RonGordon After adding both sides there was $2$ on both side which cancelled. I don't want to write it $again^2$ :). $\endgroup$ – ABC Apr 8 '13 at 14:30
  • $\begingroup$ My mistake, sorry about that. $\endgroup$ – Ron Gordon Apr 8 '13 at 14:32
  • 16
    $\begingroup$ Where is Gauss Mean Value theorem here? $\endgroup$ – Did Apr 8 '13 at 16:40
  • $\begingroup$ Aah ! Bounty really get's attention of people. I suddenly got a large attention after a month. $\endgroup$ – ABC May 2 '13 at 15:55

My try: \begin{align} \int_0^{\pi}\ln\sin\theta\,d\theta=\frac{1}{2}\int_{-\pi/2}^{\pi/2}\left(-\ln 4+\ln4\cos^2\theta\right)d\theta=-\pi\ln 2+\frac12\int_{-\pi/2}^{\pi/2}\ln(2+2\cos2\theta)d\theta=\\= -\pi\ln 2+\frac14\underbrace{\int_{-\pi}^{\pi}\left[\ln(1+e^{i\theta})+\ln(1+e^{-i\theta})\right]d\theta}_{=0\;\mathrm{by\; MVT}}=-\pi\ln 2. \end{align}

  • $\begingroup$ nice explanation................ $\endgroup$ – juantheron Oct 27 '13 at 17:41

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.