A challenging integral

[Tolaso J Kos]

Prove that:

$$\int_{0}^{\pi/2}\frac{x^2}{x^2+\ln^2 (2\cos x)}\, {\rm d}x = \frac{\pi}{8}\left ( 1-\gamma + \ln 2\pi \right )$$

[r9m]

We begin by observing that: \begin{align} \int_{-\pi/2}^{\pi/2} \left(e^{ix}+e^{-ix}\right)^s e^{iwx}\,dx&= \frac{1}{i}\int\limits_{\gamma} \left(z+z^{-1}\right)^s z^{(w-1)}\,dz\\&= \frac{1}{i}\int\limits_{\gamma} \left(z+z^{-1}\right)^s z^{(w-1)}\,dz\\&= \frac{1}{i}\int\limits_{\gamma} \left(1+z^{2}\right)^s z^{(w-s-1)}\,dz \end{align}

where, $\gamma$ is the semi-circular contour $\substack{|z| = 1\\ -\pi/2 <\arg(z) < \pi/2}$ in the positive half plane with indentations at $z = \pm i$. We complement the contour with the line joining $z = -i$ to $z = i$ except with indentation at $z = -i,0,i$.

\begin{align}\int\limits_{\gamma} \left(1+z^{2}\right)^s z^{(w-s-1)}\,dz &= e^{i\frac{\pi}{2}(w-s)}\int_{0}^{1} (1-x^2)^sx^{w-s-1}\,dx+e^{i\frac{\pi}{2}(w-s)}\int_0^{-1} (1-x^2)^sx^{w-s-1}\,dx\\&= e^{i\frac{\pi}{2}(w-s)}\int_{0}^{1} (1-x^2)^sx^{w-s-1}\,dx-e^{-i\frac{\pi}{2}(w-s)}\int_0^{1} (1-x^2)^sx^{w-s-1}\,dx\\&= 2i\sin \left(\frac{\pi (w-s)}{2}\right)\int_0^1 x^{w-s-1}(1-x^2)^s\,dx\\&= i\sin \left(\frac{\pi (w-s)}{2}\right)B\left(\frac{w-s}{2},s+1\right)\end{align}

Thus, \begin{align}\int_{-\pi/2}^{\pi/2} \left(e^{ix}+e^{-ix}\right)^s e^{iwx}\,dx &= \sin \left(\frac{\pi (w-s)}{2}\right)B\left(\frac{w-s}{2},s+1\right)\end{align}

Differentiating with respect to $w$ we have:

\begin{align}&\int_{-\pi/2}^{\pi/2} x\left(e^{ix}+e^{-ix}\right)^s e^{iwx}\,dx\\&=-i\frac{d}{dw} \sin \left(\frac{\pi (w-s)}{2}\right)B\left(\frac{w-s}{2},s+1\right) \\&= -iB\left(\frac{w-s}{2},s+1\right)\left(\frac{1}{2}\sin \left(\frac{\pi (w-s)}{2}\right)\left(\psi\left(\frac{w-s}{2}\right) - \psi\left(\frac{w+s}{2}+1\right)\right) + \frac{\pi}{2}\cos \left(\frac{\pi (w-s)}{2}\right)\right)\end{align}

Now, for the integral at hand: \begin{align}I&=\int_{0}^{\pi/2}\frac{x^2}{x^2+\log^2 (2\cos x)}\,dx \\&= \frac{1}{2}\int_{-\pi/2}^{\pi/2}\frac{x^2}{x^2+\log^2 (2\cos x)}\,dx\\&= \frac{1}{2}\mathfrak{I}\int_{-\pi/2}^{\pi/2}\frac{x}{\log (2\cos x) - ix}\,dx\\&= \frac{1}{2}\mathfrak{I}\int_{-\pi/2}^{\pi/2}\frac{x}{\log \left(1+e^{-2ix}\right)}\,dx\\&= \frac{1}{2}\mathfrak{I}\int_{-\pi/2}^{\pi/2}\int_0^1 \frac{x(1+e^{-2ix})^s}{(1+e^{-2ix}) - 1}\,ds\,dx\\&= \frac{1}{2}\mathfrak{I}\int_0^1\int_{-\pi/2}^{\pi/2} x(e^{ix}+e^{-ix})^se^{i(2-s)x}\,dx\,ds\end{align}

Setting $w = 2-s$ in the derived identity and using $\displaystyle B(1-s,1+s) = \frac{\pi s}{\sin \pi s}$ and $\psi(1-s) - \psi (s) = \pi\cot \pi s$ we simplify it to:

\begin{align}I &= -\frac{\pi}{4}\int_0^1 s(\psi (1-s) - \psi (2) - \pi \cot \pi s)\,ds \\&= -\frac{\pi}{4}\int_0^1 s(\psi (s) - \psi (2))\,ds\\&= -\frac{\pi}{4}\left(-\frac{1}{2}+\frac{\gamma}{2} - \int_0^1 \log \Gamma(s)\,ds\right)\\&= \frac{\pi}{8}\left(1-\gamma + \log (2\pi)\right)\end{align}

The evaluation of the last $\log \Gamma$ integral can be found here: Integral involving $\Gamma$ function