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

Question

Here is an interesting, albeit tough, integral I ran across. It has an interesting solution which leads me to think it is doable. But, what would be a good strategy?.

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

This looks rough. What would be a good start?. I tried various subs in order to get it into some sort of shape to use series, LaPlace, something, but made no real progress.

I even tried writing a geometric series. But that didn't really result in anything encouraging.

$$\int_{0}^{\frac{\pi}{2}}\sum_{n=0}^{\infty}(-1)^{k}\left(\frac{\ln(2\cos(x))}{x}\right)^{2k}$$

Thanks all.

Answer 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.

Answer 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 !

Answer 3

We start by observing the following: \begin{aligned} \int_{-\pi/2}^{\pi/2} \left(e^{ix}+e^{-ix}\right)^s e^{iwx}\,dx&= \frac{1}{i} \oint \limits_{\gamma} \left(z+z^{-1}\right)^s z^{(w-1)}\,dz\\ &= \frac{1}{i} \oint \limits_{\gamma} \left(z+z^{-1}\right)^s z^{(w-1)}\,dz\\ &= \frac{1}{i} \oint\limits_{\gamma} \left(1+z^{2}\right)^s z^{(w-s-1)}\,dz \end{aligned} where $\gamma$ a semicircle with in the upper positive halfplane with $\displaystyle{\substack{|z| = 1\\ -\pi/2 <\arg(z) < \pi/2}}$ recesses at the points $z = \pm i$. We complete the semicircle with a straight section that joins the points $z=-i$ and $z=i$ apart from the recesses $z=-i$,$0$, i. Then: \begin{aligned}\oint\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{aligned} Consequently $$\displaystyle{\int_{-\pi/2}^{\pi/2} \left(e^{ix}+e^{-ix}\right)^s e^{iwx}\,dx = \sin \left(\frac{\pi (w-s)}{2}\right)B\left(\frac{w-s}{2},s+1\right)}$$ Producing as to w we have \begin{aligned}&\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{aligned} So for the original integral we have: \begin{aligned}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{aligned} Setting w=2-s and using formulas $\displaystyle{B(1-s,1+s) = \frac{\pi s}{\sin \pi s}}$ and $\displaystyle{\psi(1-s) - \psi (s) = \pi\cot \pi s}$ the result is simplified in \begin{aligned}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{aligned}