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It would take too long to explain the context reasonably well - but in short, this integral, or rather its equivalent $$\int\frac{x\cos x\,dx}{\sqrt{1-k^2\sin^2x}},\qquad 0<k<1$$ is related to generating function of a canonical transformation between two distinguished sets of local Darboux coordinates on a certain $SL(2,\mathbb{C})$ character variety.

I would be grateful for any help with the calculation of this integral. It would be especially nice to hear from people industrially guessing closed form expressions for integrals which look much less computable. The answer will be awarded a bounty, unless I manage to find it myself first.

P.S. I have reasons to believe that

  1. The answer can be found in a closed form.

  2. Moreover, I expect it to be given by a linear combination of dilogarithms $\operatorname{Li}_2(\cdot)$ with arguments expressed in terms of elementary functions (roughly, square roots and trigonometric functions).

P.P.S. I intentionally do not tag my question as "integration", "calculus" etc, as these happen to be ignored tags of some highly-qualified participants. Please don't modify this!

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  • $\begingroup$ @Babak S. Although it looks a bit like elliptic integral, I am more or less sure they will not be of any help here. But it makes no harm to include the corresponding tag anyway. $\endgroup$ – Start wearing purple Nov 6 '13 at 16:01
  • $\begingroup$ Sorry for that. Apology. @O.L. $\endgroup$ – mrs Nov 6 '13 at 16:02
  • $\begingroup$ Up to a first integration by parts, this is equivalent to find a closed form for $\int\frac{\arcsin(k\sin u)}{k}du$, i.e. $\int\sum_{j=0}^{+\infty}\frac{\binom{2j}{j}}{(2j+1)4^j}k^{2j+1}\sin^{2j+1}(x)dx$. $\endgroup$ – Jack D'Aurizio Nov 11 '13 at 1:01
  • $\begingroup$ Notice that if we take the definite integral between $0$ and $\pi$ what we get is exactly $\operatorname{Li}_2(k)-\operatorname{Li}_2(-k)$. $\endgroup$ – Jack D'Aurizio Nov 11 '13 at 1:06
  • $\begingroup$ And if we take the integral between $0$ and $\pi/2$ we get $\operatorname{Li}_2(k)-\frac{1}{4}\operatorname{Li}_2(k^2)$. $\endgroup$ – Jack D'Aurizio Nov 11 '13 at 1:12
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Let us denote $$ \mathcal{I}\left(k,x\right)=\int_0^x \eta\left(y\right)dy, \qquad \eta\left(y\right)=\arcsin\left(k\sin y\right).$$ It turned out that this integral can be computed in a rather simple way. Namely, let us make the following observations:

  1. We have $$ \eta\left(x\right)=\frac{1}{2i}\left[\ln\left(1+k\, e^{i\left(x+\eta(x)\right)}\right)-\ln\left(1+k\, e^{-i\left(x+\eta(x)\right)}\right)\right]$$
  2. Also, \begin{align*} 2\left[1+\frac{k\,\cos x}{\sqrt{1-k^2\sin^2x}}\right]\times\frac{1}{2i}\left[\ln\left(1+k\, e^{i\left(x+\eta(x)\right)}\right)-\ln\left(1+k\, e^{-i\left(x+\eta(x)\right)}\right)\right]=\\ =\frac{d}{dx}\left[\operatorname{Li}_2\left(-k\,e^{i\left(x+\eta(x)\right)}\right)+ \operatorname{Li}_2\left(-k\,e^{-i\left(x+\eta(x)\right)}\right)\right] \end{align*}
  3. And finally: \begin{align*} \eta'\left(x\right)=\frac{k\,\cos x}{\sqrt{1-k^2\sin^2x}}. \end{align*}

Combining these three formulas, we get \begin{align*}\boxed{ \;2\mathcal{I}\left(k,x\right)=\operatorname{Li}_2\left(-k\,e^{i\left(x+\eta(x)\right)}\right)+ \operatorname{Li}_2\left(-k\,e^{-i\left(x+\eta(x)\right)}\right)-2\operatorname{Li}_2\left(-k\right)-\eta^2\left(x\right)\;} \end{align*}

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