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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.

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That geometric series does not converge over the whole of the range of integration, so it is not surprising that it didn't produce encouragement. $$ \lim_{x\to\pi/2}\left(\frac{\log(2\cos(x))}{x}\right)^2=+\infty $$ –  robjohn Aug 4 '12 at 14:19
    
Yes, robjohn, that is obvious now. I was just grabbing at straws. Perhpas digamma can be implemented in some fashion. That $\ln(2\pi)$ and $\gamma$ may suggest so. Just a thought. –  Cody Aug 4 '12 at 14:21
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I strongly suspect that this is closely related to the identity $$ \binom{\alpha}{\omega} := \frac{\Gamma(1+\alpha)}{\Gamma(1+\alpha-\omega)\Gamma(1+\omega)} = \frac{1}{2\pi}\int_{-\pi}^{\pi}(1+e^{i\theta})^{\alpha}e^{-i\omega \theta}\;d\theta,$$ which holds for all $\alpha \geq 0$ and $\omega \in \mathbb{R}$, though I still cannot figure how it is related. –  sos440 Aug 4 '12 at 16:56
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Apparently you're onto something, SOS. This is mentioned in one of the papers Raymond posted links to. –  Cody Aug 4 '12 at 17:29
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2 Answers

up vote 16 down vote accepted

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.

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+1: Interesting way too! Thanks for the effort sos440! –  Raymond Manzoni Aug 4 '12 at 22:18
    
Man, SOS, you're amazing!!!. :):) –  Cody Aug 4 '12 at 22:44
    
SOS, may I ask, what are the 'simple calculations' that show the first line after the identity at (1)?. I can't believe it, but I can actually follow the rest of it, I think. :) Also when you differentiate, I assume it is w.r.t $\omega$?. –  Cody Aug 4 '12 at 23:08
    
@Cody, the easiest way I think to prove this is to use the Euler's identity. Then $$ 1 + e^{2ix} = 2 \left( \frac{e^{ix} + e^{-ix}}{2} \right) e^{ix} = 2 \cos x e^{ix}.$$ Or maybe you can prove it geometrically by considering the unit circle centered at $1$. Also, yes, the differentiation is taken w.r.t. $\omega$. –  sos440 Aug 4 '12 at 23:29
    
Thanks a lot. wonderful and ingenious proof –  Cody Aug 4 '12 at 23:40
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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) :

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Wow, I see now why I did not even know where to begin :) Thanks for all of the articles, Raymond. –  Cody Aug 4 '12 at 17:28
    
@Cody: Glad you found them interesting too! I discovered this 'mine' while searching 'log-sine' (log-cosine here) integrals. –  Raymond Manzoni Aug 4 '12 at 18:47
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