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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.

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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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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.

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**

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

$\hspace{4.5cm}$

$$

\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}

$$

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 $.

$$I=\displaystyle \int_0^{\pi} \ln \left( \sin \theta \right)\cdot d\theta$$

$$I=2\times \displaystyle \int_0^{\pi/2} \ln \left( \sin \theta \right) \cdot d\theta$$

$$I=2\times \displaystyle \int_0^{\pi/2} \ln \left( \cos \theta \right) \cdot d\theta$$

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,

$$I=-2\int_0^{\pi/2}\ln2\cdot d\theta$$

$$I=-{\pi\ln2}$$

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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