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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.

  • 1
    Apparently you're onto something, SOS. This is mentioned in one of the papers Raymond posted links to.2012-08-04

2 Answers 2

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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.

  • 0
    @vesszabo, it is the *[digamma function](http://en.wikipedia.org/wiki/Digamma_function)*.2012-08-08
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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) :

  • 0
    @Nemo: I don't remember the exact reason of this inclusion but $(1.10)$ for example contains $\int_0^1 \ln\Gamma(x)\, dx$ that matters in Sangchul Lee's answer. I notice too generalizations of $\int_0^1 t\psi(t) dt$ that could be of interest.2017-10-12