Wednesday, February 1, 2017

real analysis - integral $ int_0^{frac{pi}{3}}mathrm{ln}left(frac{mathrm{sin}(x)}{mathrm{sin}(x+frac{pi}{3})}right) mathrm{d}x$




We want to evaluate



$ \displaystyle \int_0^{\frac{\pi}{3}}\mathrm{ln}\left(\frac{\mathrm{sin}(x)}{\mathrm{sin}(x+\frac{\pi}{3})}\right)\ \mathrm{d}x$.



We tried contour integration which was not helpful. Then we started trying to use symmetries, this also doesn't work.



Our tries


Answer



Here is an approach that uses real methods only.




\begin{align*}
I &= \int_0^{\pi/3} \ln \left (\frac{\sin x}{\sin \left (x + \frac{\pi}{3} \right )} \right ) \, dx\\
&= \int_0^{\pi/3} \ln (\sin x) \, dx - \int_0^{\pi/3} \ln \left (\sin \left (x + \frac{\pi}{3} \right ) \right ) \, dx.
\end{align*}
If in the second integral appearing on the right a substitution of $x \mapsto x - \frac{\pi}{3}$ is enforced, one has
\begin{align*}
I &= \int_0^{\pi/3} \ln (\sin x) \, dx - \int_{\pi/3}^{2\pi/3} \ln (\sin x) \, dx\\
&= 2 \int_0^{\pi/3} \ln (\sin x) \, dx - \int_0^{2\pi/3} \ln (\sin x) \, dx.\tag1
\end{align*}




Now consider the integral
$$I(\alpha) = \int_0^\alpha \ln (\sin x) \, dx, \quad 0 < \alpha < \pi.\tag2$$
Taking advantage of the well-known identity
$$\ln (\sin x) = -\ln 2 - \sum_{k = 1}^\infty \frac{\cos (2kx)}{k}, \quad 0 < x < \pi,$$
substituting this result into (2), after interchanging the summation with the integration before integrating one finds
$$I(\alpha) = -\alpha \ln 2 - \frac{1}{2} \sum_{k = 1}^\infty \frac{\sin (2k \alpha)}{k^2} = -\alpha \ln 2 - \frac{1}{2} \text{Cl}_2 (\alpha).$$
Here $\text{Cl}_2 (\varphi)$ denotes the Clausen function of order two.



In terms of the Clausen function of order two the integral in (1) can be written as

$$I = -\text{Cl}_2 \left (\frac{2\pi}{3} \right ) + \frac{1}{2} \text{Cl}_2 \left (\frac{4\pi}{3} \right ).\tag3$$
From the duplication formula for the Clausen function of order two, namely
$$\text{Cl}_2 (2\theta) = 2 \text{Cl}_2 (\theta) - 2 \text{Cl}_2 (\pi - \theta), \quad 0 < \theta < \pi,$$
if we set $\theta = 2\pi/3$ in the above duplication formula, as
$$\text{Cl}_2 \left (\frac{4\pi}{3} \right ) = 2 \text{Cl}_2 \left (\frac{2\pi}{3} \right ) - 2 \text{Cl}_2 \left (\frac{\pi}{3} \right ),$$
the expression for our integral in (3) can be expressed more simply as
$$\int_0^{\pi/3} \ln \left (\frac{\sin x}{\sin \left (x + \frac{\pi}{3} \right )} \right ) \, dx = -\text{Cl}_2 \left (\frac{\pi}{3} \right ).$$


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