Let f : 0 , 2 → R be defined as f x = log 2 1 + tan π x 4 . Then, lim n → ∞ 2 n f 1 n…

Given $f\left(x\right)={\log }_2\left(1+\tan \left(\frac{\pi \mathrm{x}}{4}\right)\right)=\frac{{\log }_e\left(1+\tan \left(\frac{\pi \mathrm{x}}{4}\right)\right)}{{\log }_{e}2}$

$E=2\underset{n\to \infty }{\lim }\sum _{r=1}^n\frac{1}{n}f\left(\frac{r}{n}\right)$

$E=\frac{2}{{\log }_{e}2}{\int }_0^1{\log }_e\left(1+\tan \frac{\pi x}{4}\right)dx$   $...\left(\mathrm{i}\right)$

Replacing $x\to 1-x$

$E=\frac{2}{{\log }_{e}2}{\int }_0^1{\log }_e\left(1+\tan \frac{\pi }{4}\left(1-x\right)\right)dx$

$E=\frac{2}{{\log }_{e}2}{\int }_0^1{\log }_e\left(1+\tan \left(\frac{\pi }{4}-\frac{\pi }{4}x\right)\right)dx$

$E=\frac{2}{{\log }_{e}2}{\int }_0^1{\log }_e\left(1+\frac{1-\tan \frac{\pi }{4}x}{1+\tan \frac{\pi }{4}x}\right)dx$

$E=\frac{2}{{\log }_{e}2}{\int }_0^1{\log }_e\left(\frac{2}{1+\tan \frac{\pi x}{4}}\right)dx$

$E=\frac{2}{{\log }_{e}2}{\int }_0^1\left({\log }_{e}2-{\log }_e\left(1+\tan \frac{\pi x}{4}\right)\right)dx$    $...\left(\mathrm{ii}\right)$

Adding an equation $\left(\mathrm{i}\right) \& \left(\mathrm{ii}\right)$ we get, 

$E=1$