33
$\begingroup$

Show that

$$\begin{aligned} \int_0^{\pi/3} \log^2 \left(2\sin \frac{x}{2} \right)dx &= \frac{7\pi^3}{108} \\ \int_0^{\pi/3}x\log^2 \left(2\sin\frac{x}{2} \right)dx &= \frac{17\pi^4}{6480}\end{aligned}$$

  • I can solve $\displaystyle \int_0^\pi \log^2 \left(2\sin \frac{x}{2} \right)dx $ but I don't know what to do if the limits are from $0$ to $\pi/3$.
  • I have no idea what to do if the integrand contains an $x$.
  • I feel that the Polylogarithm function will be involved however I don't know how it can be implemented here.

It would be really great if someone could take the initiative to prove these.

$\endgroup$
  • 9
    $\begingroup$ Hello, you can see here prove , de.wikibooks.org/wiki/… $\endgroup$ – math110 Mar 27 '13 at 13:54
  • $\begingroup$ Thanks but it's not written in english. $\endgroup$ – Anthony Mar 27 '13 at 13:58
  • 1
    $\begingroup$ @Anthony. True, it's written in maths, more understandable to me than english ;-) $\endgroup$ – Jean-Claude Arbaut Mar 27 '13 at 16:57
  • $\begingroup$ By the way, the indefinite integral ${\large\int}\log^2 \left(2\sin \frac{x}{2} \right)dx$ has a closed form. $\endgroup$ – Vladimir Reshetnikov Jan 2 '16 at 20:54
21
$\begingroup$

The best way I see to do the simpler integral (the one without the $x$ in front) is to substitute $u=2 \sin{(x/2)}$ and expand the resulting integrand in a series. To wit, upon doing the substitution, we get

$$\int_0^{\pi/3} dx \: \log^2{[2 \sin{(x/2)}]} = \int_0^1 du \frac{\log^2{u}}{\sqrt{1-u^2/4}}$$

Note that

$$\frac{1}{\sqrt{1-u^2/4}} = \sum_{k=0}^{\infty} \frac{1}{2^{4 k}} \binom{2 k}{k} u^{2 k}$$

Then the integral becomes

$$\sum_{k=0}^{\infty} \frac{1}{2^{4 k}} \binom{2 k}{k} \int_0^1 du\: u^{2 k} \log^2{u}$$

The integral on the right may be done through integration by parts; the process is very interesting, but I leave it to the reader to get the nifty result that

$$\int_0^1 du\: u^{2 k} \log^2{u}=\frac{2}{(2 k+1)^3}$$

Therefore the evaluation of the integral becomes an evaluation of the following sum:

$$2 \sum_{k=0}^{\infty} \frac{1}{2^{4 k}} \binom{2 k}{k}\frac{1}{(2 k+1)^3}$$

To evaluate this sum, define

$$f(z) = 2 \sum_{k=0}^{\infty} \frac{1}{2^{4 k}} \binom{2 k}{k}\frac{z^{2 k+1}}{(2 k+1)^3}$$

The desired integral is $f(1)$. To derive an equation for $f$, recall the binomial series above:

$$\frac{d}{dz}\left [z \frac{d}{dz} \left [ z \frac{d}{dz} f(z)\right ] \right ] = 2 \sum_{k=0}^{\infty} \frac{1}{2^{4 k}} \binom{2 k}{k} z^{2 k} = \frac{2}{\sqrt{1-z^2/4}}$$

The resulting integrations are elementary except for the last one, which results in a nasty generalized hypergeometric function, something like

$$2 _4F_3\left(\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{3}{2},\frac{3}{2},\frac{3}{2},\frac{1}{4}\right)$$

which in fact does numerically check out.

$\endgroup$
  • $\begingroup$ Thanks. I like your approach. $\endgroup$ – Anthony Mar 28 '13 at 8:04
  • $\begingroup$ Hi, I am always in awe of your posts Ron. I'm reading through your answer out of curiosity and I am wondering where your identity for $ \frac{1}{\sqrt{1 - \frac{u^2}{4}}} $ comes from. $\endgroup$ – Chantry Cargill Aug 11 '14 at 3:48
  • $\begingroup$ @ChantryCargill: much appreciated! The identity comes from the Taylor series for $(1-x)^{-1/2}$ about $x=0$. $\endgroup$ – Ron Gordon Aug 11 '14 at 11:04
  • 1
    $\begingroup$ @RonGordon And we can convert this generalized hypergeometric function to a combination of elementary functions and trilogarithms as shown in this answer and comments to it, and then simplify it to get $7\pi^3/216$. $\endgroup$ – Vladimir Reshetnikov Dec 21 '15 at 21:21
  • $\begingroup$ @VladimirReshetnikov: wow, really good-looking work there. I'll need to go through it in detail so I can learn from it. $\endgroup$ – Ron Gordon Dec 21 '15 at 21:25
22
$\begingroup$

Here's another approach for evaluating the one without the $x$ in front.

First notice that it's equivalent to showing that $$\int_{0}^{\pi /6} \log^{2}(2 \sin x) \ dx = \frac{7 \pi^{3}}{216}. $$

Using the principal branch of the logarithm and assuming that $0 < x < \pi$, we have $$ \begin{align} \log(1-e^{2ix}) &= \log (e^{-ix}-e^{ix}) + \log(e^{ix}) \\ &= \log(-2i \sin x) + ix \\ &= \log(2 \sin x) - \frac{i \pi}{2} + ix. \end{align}$$

Squaring both sides and integrating, $$\int_{0}^{\pi /6} \left(\log(2 \sin x) - \frac{i \pi}{2} + ix \right)^{2} \ dx = \int_{0}^{\pi /6} \log^{2} (1-e^{2ix}) \ dx . $$

Then equating the real parts on both sides of the equation, we get

$$\begin{align} \int_{0}^{\pi /6} \log^{2}(2 \sin x) \ dx &= \int_{0}^{\pi/6} \left(x- \frac{\pi}{2} \right)^{2} \ dx + \text{Re} \int_{0}^{\pi /6} \log^{2}(1-e^{2ix}) \ dx \\ &= \frac{19 \pi^{3}}{648} +\text{Re} \int_{C} \log^{2}(1-z) \frac{dz}{2iz} \\ &=\frac{19 \pi^{3}}{648} + \frac{1}{2} \ \text{Im} \int_{C} \frac{\log^{2}(1-z)}{z} \ dz \end{align}$$

where $C$ is the portion of the unit circle from $z=1$ to $z=e^{ \pi i /3}$.

But since $\frac{\log^{2}(1-z)}{z}$ is analytic for $\text{Re}(z) <1$,

$$ \begin{align} \int_{C} \frac{\log^{2}(1-z)}{z} \ dz &= \int_{1}^{e^{\pi i /3}} \frac{\log^{2}(1-z)}{z} \ dz . \end{align} $$

Then integrating by parts twice, we get

$$ \begin{align} \text{Im} \int_{1}^{e^{\pi i /3}} \frac{\log^{2}(1-z)}{z} \ dz &= \text{Im} \ \log^{2}(1-z) \log(z) \Bigg|^{e^{\pi i /3}}_{1} + 2 \ \text{Im} \int_{1}^{e^{\pi i /3}} \frac{\log(1-z) \log (z)}{1-z} \ dz \\ &= \text{Im} \ \log^{2}(e^{-\pi i /3}) \log(e^{\pi i /3}) + 2 \ \text{Im} \ \log(1-z) \text{Li}_{2}(1-z) \Bigg|^{e^{\pi i / 3}}_{1} \\ &+ 2 \ \text{Im} \int_{1}^{e^{\pi i / 3}} \frac{\text{Li}_{2}(1-z)}{1-z} \ dz \\ &=- \frac{ \pi^3}{27} - \frac{2 \pi }{3}\text{Im} \ i \ \text{Li}_{2} (e^{- \pi i /3}) - 2 \ \text{Im} \ \text{Li}_{3}(1-z) \Bigg|^{e^{\pi i/3}}_{1} \\ &= - \frac{ \pi^3}{27} - \frac{2 \pi }{3}\text{Im} \ i \ \text{Li}_{2} (e^{- \pi i /3}) - 2 \ \text{Im} \ \text{Li}_{3}(e^{ -\pi i /3}) \\ &= - \frac{\pi^3}{27} - \frac{2 \pi }{3} \sum_{n=1}^{\infty} \frac{\cos (n \pi /3)}{n^{2}} +2 \sum_{n=1}^{\infty} \frac{\sin (n \pi /3)}{n^3}. \end{align}$$

Integrating both sides of the Fourier series $$\sum_{n=1}^{\infty} \frac{\sin (k \theta)}{k} = \frac{\pi - \theta}{2} \ , \ 0 < \theta < 2 \pi$$

we get

$$\sum_{n=1}^{\infty} \frac{\cos (k \theta)}{k^{2}} = \frac{\theta^{2}}{4} - \frac{\pi \theta}{2} + \frac{\pi^{2}}{6} .$$

And integrating a second time, $$ \sum_{n=1}^{\infty} \frac{\sin (k \theta)}{k^{3}} = \frac{\theta^{3}}{12} - \frac{\pi \theta^{2}}{4} + \frac{\pi^{2} \theta}{6}.$$

Therefore,

$$\sum_{n=1}^{\infty} \frac{\cos (n \pi /3)}{n^{2}} = \frac{\pi^{2}}{36} $$

and $$ \sum_{n=1}^{\infty} \frac{\sin (n \pi /3)}{n^{3}} = \frac{5 \pi^{3}}{162}. $$

So finally we have

$$ \begin{align} \int_{0}^{\pi /6} \log^{2}(2 \sin x) \ dx &= \frac{19 \pi^{3}}{648} + \frac{1}{2} \left[ - \frac{ \pi^{3}}{27} - \frac{2 \pi }{3} \left(\frac{\pi^{2}}{36} \right) + 2 \left( \frac{5 \pi^{3}}{162} \right) \right] \\ &= \frac{7 \pi^{3}}{216} . \end{align}$$

$\endgroup$
  • $\begingroup$ Random Variable, your solution is very nice. But in your answer and the one form de.wikibooks.org/wiki/…, complex analysis is used. I do not know if there is a way to use only real analysis to solve the problem. $\endgroup$ – xpaul Dec 7 '14 at 16:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.