Question

Let $f:(0,2)\to R$ be defined as
$f(x)={\log }_2\left(1+\tan \left(\frac{\pi x}{4}\right)\right).$
Then, ${\lim }_{n\to \infty }\frac{2}{n}\left(f\left(\frac{1}{n}\right)+f\left(\frac{2}{n}\right)+\dots +f(1)\right)$ is equal to _______ .

Answer 1

Answer: 1

$E=2{\lim }_{n\to \infty }{\sum }_{r=1}^n\frac{1}{n}f\left(\frac{r}{n}\right)$
$E=\frac{2}{\ln 2}{\int }_0^1\ln \left(1+\tan \frac{\pi x}{4}\right)dx$
replacing $x\to 1-x$
$E=\frac{2}{\ln 2}{\int }_0^1\ln \left(1+\tan \frac{\pi }{4}(1-x)\right)dx$
$E=\frac{2}{\ln 2}{\int }_0^1\ln \left(1+\tan \left(\frac{\pi }{4}-\frac{\pi }{4}x\right)\right)dx$
$E=\frac{2}{\ell n2}{\int }_0^1\ln \left(1+\frac{1+\tan \frac{\pi }{4}x}{1+\tan \frac{\pi }{4}x}\right)dx$
$E=\frac{2}{\ln 2}{\int }_0^1\ln \left(\frac{2}{1+\tan \frac{\pi x}{4}}\right)dx$
$E=\frac{2}{\ell n2}{\int }_0^1\left(\ln 2-\ln \left(1+\tan \frac{\pi x}{4}\right)\right)dx$
equation (i) $+$ (ii) $E=1$