- Thread starter
- #1

- Jan 17, 2013

- 1,667

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

*where*\(\displaystyle \log\)

*represents the natural logarithm*

- Thread starter ZaidAlyafey
- Start date

- Thread starter
- #1

- Jan 17, 2013

- 1,667

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

- Moderator
- #2

- Jan 26, 2012

- 995

\[\frac{1}{\pi}\int_0^{\pi}\log(\sin u)\,du.\]

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.

\[\left|\int_L f(z)\,dz\right|\leq 2\pi|e^{iz}|=2\pi e^{-y}\rightarrow 0\]

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)

- Thread starter
- #3

- Jan 17, 2013

- 1,667

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

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

- Thread starter
- #4

- Jan 17, 2013

- 1,667

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

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)

\)

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

- Moderator
- #5

- Jan 26, 2012

- 995

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

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

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

- Thread starter
- #6

- Jan 17, 2013

- 1,667

\(\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*}

\)

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

- Admin
- #7

Nice use of the property:

\(\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*}

\)

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

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