Limits and Integrals (hum, yummy)

Take $g : I =[1, \infty) \to \mathbb{R},\text{ such that } g(t) = \frac{\pi t}{2t + 1}$ . This function is a strictly function because $g'(t) = \frac{\pi}{(2t +1)^2} > 0,\space \forall t \in (1,\infty)$ , and $g(1) = \frac{\pi}{3}$ and $\displaystyle \lim_{t\to \infty} g(t) = \frac{\pi}{2}$ .

Since $\tan(x) > x , \space \forall x \in (\frac{\pi}{4},\frac{\pi}{2})$ we have: $\displaystyle\lim_{x\to \infty} \int_{1}^x \frac{1}{t}\tan\left(\frac{\pi t}{2t +1}\right) dt \ge \displaystyle\lim_{x\to \infty} \int_{1}^x \frac{1}{t} \cdot \frac{\pi t}{2t + 1} dt \ge \displaystyle\lim_{x\to \infty} \int_{1}^x \frac{\pi}{2t + 1} dt =$ $= \frac{\pi}{2} \cdot \left[\displaystyle\lim_{x\to \infty} \ln(2x + 1) - \ln(3)\right] = \infty$ Thus we have $\displaystyle\lim_{x\to \infty} \frac{1}{x} \int_{1}^x \frac{1}{t}\tan\left(\frac{\pi t}{2t +1}\right) dt = \frac{\infty}{\infty}$ so that we can apply L'Hopital's rule and the fundamental theorem of calculus (version 1), obtaining $\displaystyle\lim_{x\to \infty} \frac{1}{x} \int_{1}^x \frac{1}{t}\tan\left(\frac{\pi t}{2t +1}\right) dt = \lim_{x \to \infty} \frac{1}{x} \tan\left(\frac{\pi x}{2x + 1}\right) =$ $= \displaystyle\lim_{x\to \infty} \frac{\frac{1}{x}}{\cos\left(\frac{\pi x}{2x + 1}\right)} \times \lim_{x\to \infty} \sin\left(\frac{\pi x}{2x + 1}\right) = \frac{0}{0}.$ Now, it's sufficient that there exists the first limit above because the second limit above is $1$ , and applying L'Hopital's rule again for the first limit we get $\displaystyle\lim_{x\to \infty} \frac{\frac{1}{x}}{\cos\left(\frac{\pi x}{2x + 1}\right)} = \lim_{x\to \infty} \frac{\frac{-1}{x^2}}{- \sin\left(\frac{\pi x}{2x + 1}\right) \cdot \frac{\pi}{(2x +1)^2}} = \lim_{x\to \infty} \frac{(2x +1)^2}{\pi x^2} = \frac{4}{\pi}.$ Therefore, $\displaystyle\lim_{x\to \infty} \frac{1}{x} \int_{1}^x \frac{1}{t}\tan\left(\frac{\pi t}{2t +1}\right) dt = \frac{4}{\pi}$

Suppose that $f(t)$ is a continuous function on $[1,\infty)$ such that $\lim_{t\to\infty}f(t) = L$ . Then $\frac{1}{x}\int_1^x f(t)\,dt - L \; = \; \frac{1}{x}\int_1^x \big[f(t) - L\big]\,dt - \frac{L}{x}$ Given $\epsilon > 0$ find $X > 1$ such that $|f(t)-L| < \epsilon$ for all $t > X$ . Then $\left|\frac{1}{x}\int_1^x f(t)\,dt - L\right| \; \le \; \frac{1}{x}\left|\int_1^X \big[f(t) - L\big]\,dt\right| + \epsilon\frac{x-X}{x} + \frac{|L|}{x} \; \le \; \epsilon + \frac{1}{x}\left((X-1)\sup_{1 \le t \le X}|f(t)-L| + |L|\right)$ for all $x > X$ , so we can find $Y > X$ such that $\left|\frac{1}{x}\int_1^x f(t)\,dt - L\right| \; < \; 2\epsilon$ for all $x > Y$ . In other words, we have shown that $\lim_{x \to \infty}\frac{1}{x}\int_1^x f(t)\,dt \; = \; L$ In this case we have $f(t) \; = \; \frac{1}{t}\tan\left(\frac{\pi t}{2t+1}\right) \; = \; \frac{1}{t}\cot\left(\frac{\pi}{2(2t+1)}\right) \;=\; \frac{\cos\left(\frac{\pi}{2(2t+1)}\right)}{g\left(\frac{\pi}{2(2t+1)}\right)} \times \frac{2(2t+1)}{\pi t}$ Where $g(x)=\frac{\sin x}{x}$ . Since $g(x) \to 1$ as $x \to 0$ , we deduce that $f(t) \to \frac{4}{\pi}$ as $t \to \infty$ . Thus the solution to this problem is $\tfrac{4}{\pi} = \boxed{1.273}$ .