A limit evaluating to $2 K$ (Catalan's constant)

Solution 1:

Denote $\displaystyle u_j=\cos \frac{x}{2^j}$, and rewrite the integrand as \begin{align}\sum_j\frac{u_j^{-1}}{\prod_{k\neq j}\left(u_j-u_k\right)}&=\frac{1}{\prod_{m}u_m} \sum_j \prod_{k\neq j}\frac{0-u_k}{u_j-u_k}=\frac{1}{\prod_{m}u_m}, \end{align} where at the last step we used the Lagrange interpolation formula (see for example the answers to this question).

The product in the denominator is telescoping: using the double sine formula, we obtain $$\sin \frac{x}{2^{2n+1}}{\prod_{m}u_m}=\frac{1}{2^{2n+1}}\sin x.$$

Therefore, the limit reduces to $$\lim_{n\to\infty}\int_0^{{\pi}/{2}}\frac{2^{2n+1}\sin \frac{x}{2^{2n+1}}}{\sin x}dx= \int_0^{{\pi}/{2}}\frac{xdx}{\sin x}=2K,$$ where the last equality is essentially the formula (40) from Wolfram page on Catalan's constant.