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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?

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  • $\begingroup$ 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, 2013 at 11:56
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    $\begingroup$ Similar question here: math.stackexchange.com/questions/37829/… $\endgroup$
    – Cocopuffs
    Apr 8, 2013 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. $\endgroup$ Apr 8, 2013 at 12:15
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    $\begingroup$ 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, 2013 at 13:24
  • $\begingroup$ @muzzlator yes that I am. $\endgroup$ Apr 8, 2013 at 14:48

5 Answers 5

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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

$\mathrm{Re}(\log(z))=\log(|z|)=\log\left(\sqrt{2-2\cos(x)}\right)$

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

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    $\begingroup$ (+1.5) the picture itself should worth an extra 1/2 upvote ;-p $\endgroup$ May 2, 2013 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, 2014 at 19:45
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    $\begingroup$ @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, 2014 at 1:06
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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.

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  • $\begingroup$ all right ... thanks!! $\endgroup$ May 8, 2013 at 20:42
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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 $.

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  • $\begingroup$ very interesting!! i knew the identity $\endgroup$ May 8, 2013 at 20:38
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$$I=\displaystyle \int_0^{\pi} \ln \left( \sin \theta \right)\cdot d\theta$$ $$I=2\times \displaystyle \int_0^{\pi/2} \ln \left( \sin \theta \right) \cdot d\theta$$ $$I=2\times \displaystyle \int_0^{\pi/2} \ln \left( \cos \theta \right) \cdot d\theta$$ 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, $$I=-2\int_0^{\pi/2}\ln2\cdot d\theta$$ $$I=-{\pi\ln2}$$

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  • $\begingroup$ @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, 2013 at 14:30
  • $\begingroup$ My mistake, sorry about that. $\endgroup$
    – Ron Gordon
    Apr 8, 2013 at 14:32
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    $\begingroup$ Where is Gauss Mean Value theorem here? $\endgroup$
    – Did
    Apr 8, 2013 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, 2013 at 15:55
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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}

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  • $\begingroup$ nice explanation................ $\endgroup$
    – juantheron
    Oct 27, 2013 at 17:41

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