Interesting integral

Question

Numerical evidence shows the validity of the following identity $$\int\limits_0^z\frac{xdx}{\sin{x}\sqrt{\sin^2{z}-\sin^2{x}}}=\frac{\pi}{4\sin{z}}\ln{\frac{1+\sin{z}}{1-\sin{z}}},\tag{1}$$ if $0< z< \pi/2$. How can it be proved? An indirect proof can be found in the paper http://link.springer.com/article/10.1134%2FS1547477113010044 (Potential of multiphoton exchange in the scattering of light charged particles of a heavy target, by Yu.M. Bystritskiy, E.A. Kuraev and M.G. Shatnev). Formula (1) is equivalent to (12) from the paper where it is called "the marvelous identity".

Answer 1

Actually, I now think that the easiest method is to do this: Write $k=\sin z$, so that $|k|<1$, and make the substitution $x = \arcsin(k\sin\theta)$, where $0\le \theta\le \frac\pi2$. The integral becomes $$ \int_0^{\pi/2} \frac{\arcsin(k\sin\theta)}{k\sin\theta \,\,(1-k^2\sin^2\theta)^{(1/2)}}\ \mathrm{d}\theta = \int_0^{\pi/2} \sum_{n=0}^{\infty} c_n\,k^{2n}\sin^{2n}\theta\,\mathrm{d}\theta, $$ where the numbers $c_n$ are the coefficients in the even power series $$ \frac{\arcsin(t)}{t \,(1{-}t^2)^{(1/2)}} = \sum_{n=0}^{\infty} c_n t^{2n}, $$ which are easily calculated to be $$ c_n = \frac{2^{2n} (n!)^2}{(2n{+}1)!}. $$ Combining this with the well-known formula $$ \int_0^{\pi/2} \sin^{2n}\theta\,\mathrm{d}\theta = \frac{\pi}{2^{2n+1}}\ {{2n}\choose{n}}, $$ one obtains $$ \int_0^{\pi/2} \frac{\arcsin(k\sin\theta)}{k\sin\theta (1-k^2\sin^2\theta)^{(1/2)}}\ \mathrm{d}\theta = \pi \sum_{n=0}^{\infty} \frac{k^{2n}}{(4n{+}2)} = \frac{\pi}2 \sum_{n=0}^{\infty} \frac{\sin^{2n}z}{(2n{+}1)}\ . $$ The rest should be clear.

Answer 2

This integral is due to Lobachevskii. He gave it in more general form as follows which can be found in his work "Application of imaginary geometry to certain integrals" (1836). Also see equation 3.842.7 in Gradsteyn and Ryzhik. The integral in OP's post follows from this formula in the limit $~\alpha\to i\infty$.