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I am trying to prove $$ \int_{\pi/6}^{\pi/2}\bigg[\Re\big(\text{Li}_2(4\sin^2\theta)\big) +\text{Li}_2\bigg(\frac{1}{4\sin^2\theta}\bigg) \bigg]d\theta=\frac{5\pi^3}{54}. $$ Clearly, this closed form result is very nice. I am very rusty with working with dilogarithm integrals and am not sure where to start this. Some information that may help is, note the dilogarithm function is given by $$ Li_2(z)=\sum_{n=1}^\infty \frac{z^n}{n^2}. $$ This integral is related to the dilogarithm representation $$ \frac{1}{\pi}\int_0^\pi \Re\big(Li_2(4 \sin^2 \theta)\big)d\theta=\zeta(2)=\frac{\pi^2}{6} $$ which is strongly related to the log-sine integrals I have been posting.

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    $\begingroup$ See DLMF 25.12.4 for a possible way to simplify the integrand. $\endgroup$ – Chen Wang Apr 13 '14 at 22:40
  • $\begingroup$ @ChenWang Nice, very nice. Thanks. But do you know how to prove that? $\endgroup$ – Jeff Faraci Apr 15 '14 at 3:11
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begin with the identity:

$$Li_{2}(z)+Li_{2}(1/z)=\frac{-\pi^{2}}{6}-1/2log^{2}(-z)$$

this leads to:

$$\frac{-\pi^{2}}{6}\int_{\pi/6}^{\pi/2}d\theta -1/2\int_{\pi/6}^{\pi/2}log^{2}(-4\sin^{2}\theta)d\theta$$

Of course, the first integral is very easy and evaluates to $\pi^{3}/18$.

expanding the integral gives: $$-\pi^{3}/18-1/2\int_{\pi/6}^{\pi/2}\left[4 \log^{2}(\sin\theta)+8 \log(2)log(\sin\theta) +4 \log^{2}(2)-\pi^{2}+4\pi i \log(2)+4\pi i \log(\sin\theta)\right]d\theta...(2)$$

what I do is integrate from $0, \pi/6$ and from $0,\pi/2$, then subtract them in order to obtain the result.

So, there are several log-trig integrals to evaluate:

$$\int_{0}^{\pi/6}\log^{2}(\sin\theta)d\theta$$

Let $x=\sin\theta$:

$$\int_{0}^{1/2}\frac{\log(x)}{\sqrt{1-x^{2}}}dx$$

$$1/2\log^{2}(x)\sum_{k=0}^{\infty}\binom{-1/2}{k}(-1)^{k}x^{2k}dx$$

$$=\sum_{k=0}^{\infty}\frac{(2k)!}{4^{k}(k!)^{2}}\int_{0}^{1/2}x^{2k}\log^{2}(x)dx$$

$$=\frac{\log^{2}(2)}{2}\sum_{k=0}^{\infty}\frac{(2k)!}{(2k+1)(k!)^{2}16^{k}}+\log(2)\sum_{k=0}^{\infty}\frac{(2k)!}{(2k+1)^{2}(k!)^{2}16^{k}}$$ $$+\sum_{k=0}^{\infty}\frac{(2k)!}{(2k+1)^{3}(k!)^{2}16^{k}}.......(1)$$

to evaluate the series, use the famous identity:

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

to evaluate the left series, integrate this identity, divide by x, and let $x=1/4$

Giving: $$\sum_{k=0}^{\infty}\frac{(2k)!}{(2k+1)(k!)^{2}16^{k}}=\frac{\sin^{-1}(1/2)}{1/2}=\frac{\pi}{3}$$

the middle series requires a little more.

we have to integrate $$\int_{0}^{1/4}\frac{\sin^{-1}(2x)}{2x}dx$$, which comes from integrating the aforementioned identity.

let $t=2x$

$$1/2\int_{0}^{1/2}\frac{\sin^{-1}(t)}{t}dt$$

let $u=\sin^{-1}(t)$

$$1/2\int_{0}^{\pi/6}u\cot(u)du=\frac{\sin(\pi n/3)}{4n^{2}}-\frac{\pi \cos(\pi n/3)}{12n}$$

Two Clausen series to deal with:

the left one: $$\sum_{n=1}^{\infty}\frac{\sin(\pi n/3)}{n^{2}}$$

There is plenty of info on these. I can show one way to evaluate it, but to save time let's say for now that it evaluates to $$\frac{3\psi'(1/3)-\psi'(5/6)}{16\sqrt{3}}$$

the rightmost series is an already famous series for $\pi^{3}$ because:

$$\pi^{3}=\frac{216}{7}\sum_{k=0}^{\infty}\frac{(2k)!}{(2k+1)^{3}(k!)^{2}16^{k}}$$

which means we have $$\frac{7}{216}\pi^{3}$$.

Putting all the series in (1) together results in:

$$\int_{0}^{\frac{\pi}{6}}\log^{2}(\sin\theta)d\theta=\frac{7\pi^{3}}{216}+\frac{\pi}{6}\log^{2}(2)+\log(2)\left(\frac{\sqrt{3}}{16}\psi'(1/3)-\frac{1}{16\sqrt{3}}\psi'(5/6)\right)$$

Now, $$\int_{0}^{\pi/6}\log(\sin\theta)d\theta$$ can be done along the same lines. But, to save time and space let's say it evaluates to $$-\frac{\pi}{6}\log(2)-\frac{3\psi'(1/3)-\psi'(5/6)}{16\sqrt{3}}$$

In order to find $$\int_{\pi/6}^{\pi/2}\log^{2}(\sin\theta)d\theta$$, we have to subtract the above result from the classic

$$\int_{0}^{\pi/2}\log^{2}(\sin\theta)d\theta=\frac{\pi^{3}}{24}+\frac{\pi}{2}log^{2}(2)$$

this gives:

$$\int_{\pi/6}^{\pi/2}\log^{2}(\sin\theta)d\theta=\frac{\pi^{3}}{108}+\frac{\pi}{3}\log^{2}(2)-\log(2)\left(\frac{3\psi'(1/3)-\psi'(5/6)}{16\sqrt{3}}\right)$$

similarly:

$$\int_{\pi/6}^{\pi/2}\log(\sin\theta)d\theta=\frac{-\pi}{3}\log(2)+\frac{3\psi'(1/3)-\psi'(5/6)}{3\sqrt{16}}$$

finally, going back to the beginning at (2) and putting it all in and evaluating the more elementary integrals in the same:

$$\frac{-\pi^{3}}{18}-2\int_{\pi/6}^{\pi/2}\log^{2}(\sin\theta)d\theta-4\log(2)\int_{\pi/6}^{\pi/2}\log(\sin\theta)d\theta-\frac{2\pi}{3}\log^{2}(2)$$ $$+\frac{\pi^{3}}{6}-\frac{2\pi^{2}}{3}\log(2)i-2\pi i\int_{\pi/6}^{\pi/2}\log(\sin\theta)d\theta$$

Sub in the results from above and use real parts:

$$\frac{\pi^{3}}{9}-2\left(\frac{\pi^{3}}{108}+\frac{\pi}{3}\log^{2}(2)-\log(2)\left(\frac{3\psi'(1/3)-\psi'(5/6)}{16\sqrt{3}}\right)\right)$$ $$-4\log(2)\left(\frac{-\pi}{3}\log(2)+\frac{3\psi'(1/3)-\psi'(5/6)}{16\sqrt{3}}\right)-\frac{2\pi}{3}\log^{2}(2)$$

All of this mess cancels except for

$$\frac{\pi^{3}}{9}-\frac{\pi^{3}}{54}=\frac{5\pi^{3}}{54}$$

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  • $\begingroup$ Nice work +1, I have plenty of literature on the Clausen sums and log sine integrals so Thanks! $\endgroup$ – Jeff Faraci May 25 '14 at 16:48

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