How to study the convergence of $\int^{\infty}_{0}\frac{2\cos(t)\sin(t)}{t}dt$?

Question

By using Taylor series, I managed to see that the $t \rightarrow 0$, $\frac{2\cos(t)\sin(t)}{t} = 1 -\frac{4t^2}{6}$ whose integral converges. As for the case when $t \rightarrow \infty$, I don't know that to do. It equals to $\int^{\infty}_{0}\frac{\sin(2t)}{t}dt$. What can I do?

Answer 1

$$\eqalign{\int_0^{\infty} \frac{\sin(2t)}{t}\; dt &= \sum_{n=0}^\infty \int_{n\pi/2}^{(n+1)\pi/2} \frac{\sin(2t)}{t}\; dt \cr &= \sum_{n=0}^\infty \int_{0}^{\pi/2} \sin(2t) \left(\frac{1}{t+n\pi/2} - \frac{1}{t+(n+1)\pi/2}\right)\cr &= \sum_{n=0}^\infty \int_0^{\pi/2} \frac{2\pi\;\sin(2t)}{n^2 \pi^2 + \ldots}}$$ Now estimate...

Answer 2

$\int^{\infty}_{0}\frac{sin(2t)}{t}dt$. exists if $\int^{\infty}_{0}\frac{sin(2t)} {t}dt$ exists

exists if $(1)\lim_{a\to0}\int^{1}_{a}\frac{sin(t)}{t}dt$ exists and $(2)\lim_{a\to\infty}\int^{b}_{1}\frac{sin(t)}{t}dt$ exists

$(1)$ on $[0;1], \sin(x)\leq x, \sin(x)/x\leq1$

$\int^{1}_{a}\frac{sin(t)}{t}dt \leq \int^{1}_{a}dt$, so $\lim_{a\to0}\int^{1}_{a}\frac{sin(t)}{t}dt$ exists

$(2)$ let $u_n = \int^{\pi(n+2)}_{\pi(n+1)}\frac{sin(2t)}{t}dt $

$\lim_{a\to\infty}\int^{b}_{1}\frac{sin(t)}{t}dt$ exists if $\sum u_n$ converges. $\sum u_n$ is an alternating series, so $\lim_{a\to\infty}\int^{b}_{1}\frac{sin(t)}{t}dt$ exists.