오늘의 계산 48 - A Nearly Trivial Integral

Problem. Prove that^[1] \begin{equation}\label{prob:wts} \int_{0}^{\frac{\pi}{2}} \frac{x^2}{x^2 + \log^2(2\cos x)} \; dx = \frac{\pi}{8}\left( 1 - \gamma + \log(2\pi) \right). \end{equation}

Proof. Here I refer to the following identity \begin{equation}\label{eq:01} \binom{\alpha}{\omega} = \frac{1}{2\pi} \int_{-\pi}^{\pi} \left(1 + e^{i\theta}\right)^{\alpha} e^{-i\omega \theta} \; d\theta, \end{equation} whose proof can be found in Today's Calculation No.23^[2]. Now let $x$ be a real number such that $|x| < \frac{\pi}{2}$. Then simple calculation shows that $$ \log\left(1+e^{2ix}\right) = \log(2\cos x) + ix \quad \Longrightarrow \quad \Im \left( \frac{-x}{\log\left(1+e^{2ix}\right)} \right) = \frac{x^2}{x^2 + \log^2(2\cos x)},$$ hence we have $$ \begin{align*}I &:= \int_{0}^{\frac{\pi}{2}} \frac{x^2}{x^2 + \log^2(2\cos x)} \; dx = -\int_{0}^{\frac{\pi}{2}} \Im \left( \frac{x}{\log\left(1+e^{2ix}\right)} \right) \; dx \\ &= -\frac{1}{8}\int_{-\pi}^{\pi} \Im \left( \frac{\theta}{\log\left(1+e^{i\theta}\right)} \right) \; d\theta = \frac{1}{8}\Re \left( \int_{-\pi}^{\pi} \frac{i\theta}{\log\left(1+e^{i\theta}\right)} \; d\theta \right). \end{align*}$$ Differentiating both sides of \eqref{eq:01} with respect to $\omega$ and plugging $\omega = 1$, we have $$ \frac{1}{2\pi} \int_{-\pi}^{\pi} (-i\theta) \left(1 + e^{i\theta}\right)^{\alpha} e^{-i\theta} \; d\theta = \alpha \left(\psi_0(\alpha) - \psi_0(2)\right). $$ Now integrating both sides with respect to $\alpha$ on $[0, 1]$, $$ \begin{align*} -\frac{1}{2\pi} \int_{-\pi}^{\pi} \frac{i\theta}{\log \left(1 + e^{i\theta}\right)} \; d\theta &= \int_{0}^{1} \alpha \left(\psi_0(\alpha) - \psi_0(2)\right) \; d\alpha \\ &= \left[ \alpha \log \Gamma (\alpha) \right]_{0}^{1} - \int_{0}^{1} \log \Gamma (\alpha) \; d\alpha - \frac{1}{2}\psi_0(2) \\ &= -\frac{1}{2}\left( 1 - \gamma + \log (2\pi) \right), \end{align*}$$ where we have used the fact that $$ \psi_0 (1+n) = -\gamma + H_n, \quad n \in \mathbb{N}$$ and $$ \begin{align*} \int_{0}^{1} \log \Gamma (\alpha) \; d\alpha & = \frac{1}{2} \int_{0}^{1} \log \left[ \Gamma (\alpha) \Gamma (1-\alpha) \right] \; d\alpha \\ &= \frac{1}{2} \int_{0}^{1} \log \left( \frac{\pi}{\sin \pi \alpha} \right) \; d\alpha \\ &= \frac{1}{2} \left( \log \pi - \int_{0}^{1} \log \sin \pi \alpha \; d\alpha \right) \\ &= \frac{1}{2} \log (2\pi). \end{align*} $$ Therefore we have \eqref{prob:wts}.

References

1. Show $\int_{0}^{\frac{\pi}{2}}\frac{x^{2}}{x^{2}+\ln^{2}(2\cos(x))}dx=\frac{\pi}{8}\left(1-\gamma+\ln(2\pi)\right)$ in "Math StackExchange"
2. 오늘의 계산 23 - Fourier Transform of the Extended Binomial Coefficient