Showing $\int\limits_0^{\pi}\log(\sin(\theta))d\theta = -\pi \log(2)$

Question

I have to show that

$$ \int_{0}^{\pi}\log(\sin(\theta))d\theta = -\pi \log(2) $$

This is a problem from the complex & real analysis qualifying exam.

This problem is maybe solvable by usual integration techniques but I strongly believe that they want us to prove it using complex analysis methods. Any approach will be appreciated.

I already tried with no luck:

$u = \sin(\theta)$
$u = \theta - \frac{\pi}{2}$
$\sin(\theta) = \frac{e^{i\theta} - e^{-i\theta}}{2i}$ and define a semi-circle that avoids $\{0\}$ and define a branch outsite that region (for example in the III or IV quadrants).

Perhaps you mean the integral $$\int\limits_0^{\pi}\log(\sin(\theta))d\theta = -\pi \log(2)$$, otherwise your conjectured solution is incorrect. — James Arathoon, Aug 03 '17 at 21:36
Approach three works by employing Euler's identiy.Then integrate over an rectangle in the upper half of the complex plane with one side equal to the interval $(0,\pi/2)$. Let the length of the sides paralell to the complex plane go to infinity. — tired, Aug 03 '17 at 22:17
@tired Can you post a detailed answer below? or you are just too tired? lol — Dr Richard Clare, Aug 03 '17 at 22:39

David Bowman · Accepted Answer · 2017-08-04T01:37:38.123

4

Set $I = \displaystyle \int_0^{\pi/2}\log(\cos(x))dx$. Then

\begin{align} I &= \displaystyle \frac{I+I}{2}\\ \\ &=\frac12\int_0^{\pi/2}\log\big(\sin(x))+\log(\cos(x)\big)dx \\ \\ &=\frac12\int_0^{\pi/2}\log\big(\sin(x)\cos(x)\big)dx\\ \\ &=\frac12\int_0^{\pi/2}\log\left(\frac12\sin(2x)\right)dx\\ \\ &=\frac\pi4 \log\left(\frac12\right) + \frac12\int_0^{\pi /2}\log(\sin(2x))dx\\ \\ &=\frac\pi4 \log\left(\frac12\right) + \frac I4 + \frac14\int_0^{\pi/2}\log(\cos(x))dx\\ \\ &=\frac\pi4 \log\left(\frac12\right) + \frac I4 + \frac I4\\ \\ &= \frac\pi4 \log\left(\frac12\right) + \frac I2. \end{align}

Hence $I = \displaystyle \frac\pi 2 \log\left(\frac12\right).$

Edit: perhaps my choice to attack the problem as I did is unclear. Here's maybe a little intuition for why the integrals for $\log\circ \sin$ and $\log \circ \cos$ should be equal on this interval. Look at the Riemann sums. Take a rectangle on the left half, while looking at $\log \circ \cos$. It will correspond exactly to a rectangle on the right half of the interval for $\log \circ \sin$ by the symmetry of $\sin$ and $\cos$ about $\displaystyle \frac\pi4$.

Here is a full calculation, with all the gory details.

Set $u= x-\displaystyle\frac\pi2$, so $\cos(x) = -\sin(u)$. Then we have $$\int_0^{\pi/2}\log(\cos(x))dx = \int_{-\pi /2}^0 \log(-\sin(u))du = \int_{-\pi/2}^0 \log(\sin(-u))du $$ $$= \int_{\pi/2}^0 \log(\sin(u))(-du) = \int_0^{\pi/2}\log(\sin(u))du$$ as desired.

edited Aug 04 '17 at 01:37

answered Aug 03 '17 at 20:49

David Bowman

5,572

1

How did you come with the idea that the integral of $\log(\cos(x))$ and $\log(\sin(x))$ are equal in $[0,\frac{\pi}{2}]$ ? – Dr Richard Clare Aug 03 '17 at 20:59
Answer confirmed by Mathematica. – David G. Stork Aug 03 '17 at 21:00
This is not complex analysis. – hamam_Abdallah Aug 03 '17 at 21:05
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@Richard use a change of variables. Note that $\ln \sin \theta=\ln \cos\left(\frac\pi2-\theta\right)$ and then use $\phi=\frac\pi2-\theta$. – pancini Aug 03 '17 at 21:15
@ElliotG Thank you so much. I had neither the heart nor the time to fix my formatting when I originally posted. – David Bowman Aug 04 '17 at 01:33
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@RichardClare See my edit for some intuition, but also see Elliot's comment. – David Bowman Aug 04 '17 at 01:38
@DavidBowman I already figured out that by my self when I was solving the integral again by myself. Thank you anyway for your answer. – Dr Richard Clare Aug 04 '17 at 01:53

Showing $\int\limits_0^{\pi}\log(\sin(\theta))d\theta = -\pi \log(2)$

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