Arctan and tan

If we define $F(a) \; =\; \int_0^{\frac12\pi}\frac{\tan^{-1}(a \tan x)}{\tan x}\,dx$ then $F(0) = 0$ and $\begin{aligned} F'(a) &= \int_0^{\frac12\pi} \frac{dx}{1 + a^2\tan^2x} \; = \; \int_0^{\frac12\pi} \frac{\sec^2x}{(1 + a^2\tan^2x)(1 + \tan^2x)}\,dx \\ &= \int_0^\infty \frac{du}{(1 + a^2u^2)(1 + u^2)} \; = \; \frac{1}{1-a^2}\int_0^\infty \left(\frac{1}{1+u^2} - \frac{a^2}{1 + a^2u^2}\right)\,du \\ &= \frac{1}{1-a^2}\Big[ \tan^{-1}u - a\tan^{-1}(au)\Big]_0^\infty \; = \; \frac{\pi}{2(1+a)} \end{aligned}$ for all $0 < a < 1$ , and hence $F(a) \; =\; \tfrac12\pi \ln(1+a) \hspace{2cm} 0 \le a < 1$ and then $S_N = \sum_{n=2}^N \int_0^{\frac12\pi}\frac{\tan^{-1}(\frac{1}{n} \tan x)}{\tan x}\,dx \; = \; \sum_{n=2}^N F\big(\tfrac{1}{n}\big) \;=\; \tfrac12\pi \sum_{n=2}^N \ln\big(1 + \tfrac{1}{n}\big) \; = \; \tfrac12\pi \ln\left(\prod_{n=2}^N \tfrac{n+1}{n}\right) \; =\; \tfrac12\pi \ln\big(\tfrac{N+1}{2}\big)$ so that $\exp(\pi^{-1}S_N) \,=\, \sqrt{\tfrac{N+1}{2}}$ . With $N = 2047$ , the answer is $\boxed{32}$ .