A classic log(sin) integral

[ZaidAlyafey]

Prove the following

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

where \(\displaystyle \log\) represents the natural logarithm

 

[Chris L T521]

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.

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.

 

[ZaidAlyafey]

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

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

 

[ZaidAlyafey]

Here is a method using real analysis


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

Now consider the following integral


\(\displaystyle

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)}

\)

\(\displaystyle

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)}

\)

\(\displaystyle

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)

\)


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

 

[Chris L T521]

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

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

Oops, yea that's it. I fixed that in the original post.

 

[ZaidAlyafey]

Here is an approach using elementary methods

\(\displaystyle
\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*}
\)

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

 

[MarkFL]

Here is an approach using elementary methods

\(\displaystyle
\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*}
\)

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

Nice use of the property:

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