Hello.

I am asked to calculate the number

$\displaystyle \[S = \int\limits_{\frac{\pi }{6}}^{\frac{\pi }{2}} {\sum\limits_{n = 1}^{ + \infty } {\frac{{\tan \left( {\frac{x}{{{2^n}}}} \right)}}{{{2^n}}}} } dx\]$

My approach was

$\displaystyle \[S = \int\limits_{\frac{\pi }{6}}^{\frac{\pi }{2}} {\sum\limits_{n = 1}^{ + \infty } {\frac{{\tan \left( {\frac{x}{{{2^n}}}} \right)}}{{{2^n}}}} } dx = \sum\limits_{n = 1}^{ + \infty } {\int\limits_{\frac{\pi }{6}}^{\frac{\pi }{2}} {\frac{{\tan \left( {\frac{x}{{{2^n}}}} \right)}}{{{2^n}}}} dx} = \sum\limits_{n = 1}^{ + \infty } {\ln \left( {\frac{{\cos \frac{\pi }{{3 \cdot {2^{n + 1}}}}}}{{\cos \frac{\pi }{{{2^{n + 1}}}}}}} \right)} \]$

which obviously leads to

$\displaystyle \[S = \ln \left( {\frac{{\prod\limits_{n = 1}^{ + \infty } {\cos \frac{\pi }{{3 \cdot {2^{n + 1}}}}} }}{{\prod\limits_{n = 1}^{ + \infty } {\cos \frac{\pi }{{{2^{n + 1}}}}} }}} \right)\]$

and I am not actually sure how to further calculate it (computer software shows that equations here are correct and I am getting closer to the right answer).

Thanks in advance.