Let f:(0,2)R be defined as f(x)=_2(1+ x4).

Question

Let $f:(0,2)	o\mathbb{R}$ be defined as $f(x)=\log_2\left(1+	an\dfrac{\pi x}{4}ight)$. Then, $\displaystyle\lim_{n	o\infty}\frac{2}{n}\left[f\left(\frac{1}{n}ight)+f\left(\frac{2}{n}ight)+\cdots+f(1)ight]$ is equal to..........

Answer: 1

Accepted Answer

$$\lim_{n\to\infty}\frac{2}{n}\sum_{r=1}^n f\left(\frac{r}{n}\right)=2\int_0^1 f(x)\,dx=2\log_2 e\int_0^1\ln\left(1+\tan\frac{\pi x}{4}\right)dx$$

Let $I=\displaystyle\int_0^1\ln\left(1+\tan\frac{\pi x}{4}\right)dx\ \ldots(i)$. Using $\int_a^bf(x)dx=\int_a^bf(a+b-x)dx$ with $x\to1-x$:
$$I=\int_0^1\ln\left(1+\tan\frac{\pi}{4}(1-x)\right)dx=\int_0^1\ln\left(1+\frac{1-\tan(\pi x/4)}{1+\tan(\pi x/4)}\right)dx=\int_0^1\ln\left(\frac{2}{1+\tan(\pi x/4)}\right)dx$$
$$=\int_0^1\ln2\,dx-\int_0^1\ln\left(1+\tan\frac{\pi x}{4}\right)dx\ \ldots(ii)$$

Adding (i) and (ii): $2I=\displaystyle\int_0^1\ln2\,dx=\ln2\ \Rightarrow\ I=\frac{\ln2}{2}$

So the limit $=2\log_2e\cdot\dfrac{\ln2}{2}=\log_2e\cdot\ln2=\dfrac{\ln2}{\ln2}\cdot 1=1$

Solutions · 1