MHB What is the elementary method used to prove the log(sin) integral?

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The integral $$\int^1_0 \log(\sin(\pi x)) \, dx$$ is proven to equal $$-\log(2)$$ using complex analysis and elementary methods. By changing variables and applying properties of logarithms and sine functions, the integral is transformed into a more manageable form. The discussion highlights the use of contour integration and periodicity to evaluate the integral, ultimately showing that the integral over the interval from 0 to π leads to the desired result. Additionally, an elementary approach using symmetry and properties of sine and cosine functions is also presented, confirming the same conclusion. The proof effectively demonstrates the relationship between logarithmic integrals and the value of $$-\log(2)$$.
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Prove the following

$$\int^1_0 \log(\sin(\pi x)) \, dx = - \log(2)$$

where $$\log$$ represents the natural logarithm (Smoking)
 
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Let $u=\pi x$. The integral turns into
\[\frac{1}{\pi}\int_0^{\pi}\log(\sin u)\,du.\]
Consider the function
\[f(z)=\log(1-e^{2iz})=\log(-2ie^{iz}\sin z)=\log(1-e^{-2y}(\cos(2x)+i\sin(2x))).\]
This function is real and negative if $y<0$ and $x=n\pi$. If we delete these half lines, we can assume that $\log$ is single-valued and analytic.
integral_contour-1.png
We now integrate over the rectangle with corners $z=0$, $z=\pi$, $z=\pi+iy$ and $z=iy$ (and then we let $y\to\infty$ as seen in figure). At the points $0$ and $\pi$, we create circular arcs of radius $\epsilon$ to avoid these points. By periodicity of the function, the integral along the vertical lines is zero. Also, note that if $L$ is the line connecting $iy$ with $\pi+iy$, then
\[\left|\int_L f(z)\,dz\right|\leq 2\pi|e^{iz}|=2\pi e^{-y}\rightarrow 0\]
as $y\rightarrow\infty$. Now, notice that the imaginary part of the logarithm is bounded. Thus, we only need to worry about the real part. Observe that
\[\left|\frac{1-e^{2iz}}{z}\right|\rightarrow 2\]
as $z\rightarrow 0$; thus the logarithm behaves like $\log\epsilon$. As $\epsilon\log\epsilon\rightarrow 0$,
\[\int_{C_{\epsilon}}f(z)\,dz\rightarrow 0\]

Therefore,
\[\frac{1}{\pi}\int_0^{\pi}\log(-2ie^{ix}\sin x)\,dx=0.\]
Now, consider the branch of the logarithm where $\log e^{ix} = ix$. Therefore, $\log(-i)=-\frac{\pi i}{2}$ and thus we see that
\[\frac{1}{\pi}\left[\pi\log 2-\frac{\pi^2 i}{2}+\int_0^{\pi}\log(\sin x)\,dx+\frac{\pi^2 i}{2}\right]=0\]
which implies that
\[\int_0^1\log(\sin\pi x)\,dx=\frac{1}{\pi}\int_0^{\pi}\log(\sin x)\,dx = -\log 2\]
and the proof is complete.

(I think this is good enough...let me know if there are any faulty assumptions on my part... XD)

EDIT #1: The diagram is incorrect (as I noticed when I pulled this from an old homework assignment in my complex analysis course; I'll update it in a few.)

EDIT #2: Updated the contour diagram.

EDIT #3: Fixed some typos.
 
Thanks Chris for the approach I always like a complex analysis approach even though we are actually using complex analysis using contours on the real axis .

I got confused in what function you are integrating ? , is it

$$f(z)=\log(1-e^{2iz})$$
 
Here is a method using real analysis $$\int^{\pi}_0 \log( \sin( x)) \, dx = 2\int^{\frac{\pi}{2}}_0 \log( \sin( x)) \, dx $$

Now consider the following integral $$

I(s)=2\int^{\frac{\pi}{2}}_0 (\sin( x))^s \, dx = \frac{\Gamma \left( \frac{s+1}{2}\right) \Gamma \left( \frac{1}{2}\right)}{\Gamma \left( \frac{s}{2}+1 \right)}

$$

$$

I'(s)= 2\int^{\frac{\pi}{2}}_0 (\sin( x))^s \log( \sin(x)) \, dx = \frac{\Gamma \left( \frac{1}{2}\right)}{2} \frac{\Gamma\left( \frac{s+1}{2}\right) \psi_0\left( \frac{s+1}{2}\right)-\Gamma\left( \frac{s+1}{2}\right) \psi_0 \left( \frac{s}{2}+1 \right)}{\Gamma \left( \frac{s}{2}+1 \right)}

$$

$$

I'(0)= 2\int^{\frac{\pi}{2}}_0 \log( \sin(x)) \, dx = \frac{\pi \left( \psi_0\left( \frac{1}{2}\right)- \psi_0 \left(1 \right) \right)}{2} =-\pi \log(2)

$$$$\int^1_0 \log (\sin(\pi x)) \, dx = \frac{2}{\pi } \int^{\frac{\pi}{2}}_0 \log( \sin(x)) \, dx = -\log(2)$$
 
ZaidAlyafey said:
Thanks Chris for the approach I always like a complex analysis approach even though we are actually using complex analysis using contours on the real axis .

I got confused in what function you are integrating ? , is it

$$f(z)=\log(1-e^{2iz})$$

Oops, yea that's it. I fixed that in the original post.
 
Here is an approach using elementary methods

$$
\begin{align*}
I=\int^{\pi}_0 \log( \sin( x)) \, dx &= 2\int^{\frac{\pi}{2}}_0 \log( \sin( x)) \, dx \\

&= \int^{\frac{\pi}{2}}_0 \log( \sin( x)) \, dx +\int^{\frac{\pi}{2}}_{0} \log( \sin( x)) \, dx \\

&= \int^{\frac{\pi}{2}}_0 \log( \sin( x)) \, dx +\int^{\frac{\pi}{2}}_{0} \log( \cos( x)) \, dx \\

&= \int^{\frac{\pi}{2}}_0 \log( \sin( x) \cos(x)) \, dx \\

&= \int^{\frac{\pi}{2}}_0 \log\left( \frac{\sin(2x)}{2} \right) \, dx \\

&= \int^{\frac{\pi}{2}}_0 \log( \sin(2x))\, dx -\frac{\pi}{2}\log(2) \\

&= \frac{I}{2} -\frac{\pi}{2}\log(2) \\

&=-\pi\log(2)

\end{align*}
[/Math]

$$\int^1_0 \log(\sin(\pi x))\, dx=\frac{I}{\pi} = -\log(2)$$
 
ZaidAlyafey said:
Here is an approach using elementary methods

$$
\begin{align*}
I=\int^{\pi}_0 \log( \sin( x)) \, dx &= 2\int^{\frac{\pi}{2}}_0 \log( \sin( x)) \, dx \\

&= \int^{\frac{\pi}{2}}_0 \log( \sin( x)) \, dx +\int^{\frac{\pi}{2}}_{0} \log( \sin( x)) \, dx \\

&= \int^{\frac{\pi}{2}}_0 \log( \sin( x)) \, dx +\int^{\frac{\pi}{2}}_{0} \log( \cos( x)) \, dx \\

&= \int^{\frac{\pi}{2}}_0 \log( \sin( x) \cos(x)) \, dx \\

&= \int^{\frac{\pi}{2}}_0 \log\left( \frac{\sin(2x)}{2} \right) \, dx \\

&= \int^{\frac{\pi}{2}}_0 \log( \sin(2x))\, dx -\frac{\pi}{2}\log(2) \\

&= \frac{I}{2} -\frac{\pi}{2}\log(2) \\

&=-\pi\log(2)

\end{align*}
$$

$$\int^1_0 \log(\sin(\pi x))\, dx=\frac{I}{\pi} = -\log(2)$$

(Yes) Nice use of the property:

$$\int_0^a f(x)\,dx=\int_0^a f(a-x)\,dx$$
 

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