Question

Let $f:(0,2) \rightarrow R$ be defined as
$f(x)=\log _{2}\left(1+\tan \left(\frac{\pi x}{4}\right)\right) .$
Then, $\lim _{n \rightarrow \infty} \frac{2}{n}\left(f\left(\frac{1}{n}\right)+f\left(\frac{2}{n}\right)+\ldots+f(1)\right)$ is equal to _______ .

Answer 1

$E =2 \lim _{n \rightarrow \infty} \sum_{ r =1}^{ n } \frac{1}{ n } f \left(\frac{ r }{ n }\right)$
$E =\frac{2}{\ln 2} \int_{0}^{1} \ln \left(1+\tan \frac{\pi x }{4}\right)\,d x$
replacing $x \rightarrow 1- x$
$E =\frac{2}{\ln 2} \int_{0}^{1} \ln \left(1+\tan \frac{\pi}{4}(1- x )\right)\,d x$
$E =\frac{2}{\ln 2} \int_{0}^{1} \ln \left(1+\tan \left(\frac{\pi}{4}-\frac{\pi}{4} x \right)\right)\,d x$
$E =\frac{2}{\ell n 2} \int_{0}^{1} \ln \left(1+\frac{1+\tan \frac{\pi}{4} x }{1+\tan \frac{\pi}{4} x }\right)\,d x$
$E =\frac{2}{\ln 2} \int_{0}^{1} \ln \left(\frac{2}{1+\tan \frac{\pi x }{4}}\right)\,dx$
$E =\frac{2}{\ell n 2} \int_{0}^{1}\left(\ln 2-\ln \left(1+\tan \frac{\pi x }{4}\right)\right)\,d x$
equation (i) $+$ (ii) $E =1$