Show $\int_{0}^{\frac{\pi}{2}}\frac{x^{2}}{x^{2}+\ln^{2}(2\cos(x))}dx=\frac{\pi}{8}\left(1-\gamma+\ln(2\pi)\right)$

Solution 1:

In addition to the nice set of references by Raymond Manzoni, here is my proof of the identity. Frankly, I have not seen these references yet, thus I am not sure if this already appears in one of them.

Here I refer to the following identity

$$ \binom{\alpha}{\omega} = \frac{1}{2\pi} \int_{-\pi}^{\pi} \left(1 + e^{i\theta}\right)^{\alpha} e^{-i\omega \theta} \; d\theta, \ \cdots \ (1) $$

whose proof can be found in my blog post.

Now let $x$ be a real number such that $|x| < \frac{\pi}{2}$. Then simple calculation shows that

$$ \log\left(1+e^{2ix}\right) = \log(2\cos x) + ix \quad \Longleftrightarrow \quad \Im \left( \frac{-x}{\log\left(1+e^{2ix}\right)} \right) = \frac{x^2}{x^2 + \log^2(2\cos x)},$$

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 $(1)$ 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 the desired result.

Solution 2:

To learn more about this very interesting integral I'll just provide the interesting links (the story itself is interesting since it was an experimental discovery first from Glasser and Oloa) :

• page 28 of Bailey & Borwein 'Computer-Assisted Discovery and Proof'
• Oloa's paper of 2007 'Some Euler-type integrals and a new rational series for Euler’s constant' at google or Les Mathématiques-net (or in the link 'EULER_TYPE_INTEGRALS_2007.pdf' here)
• Glasser & Mana 'The Laplace transform of the psi function'
• Dixit 2009 'The Laplace transform of the psi function'
• Amdeberhan & Moll 2010 'The Laplace transform of the digamma function: an integral due to Glasser, Manna and Oloa
• Vildanov 2010 'Generalization of the Glasser - Manna - Oloa integral and some new integrals of similar type'
• Coffey 2012 'Certain logarithmic integrals, including solution of Monthly problem #tbd, zeta values, and expressions for the Stieltjes constants'

• Weisstein has references too (see eq. (24))

  Fine reading !