Show that $\int\limits_2^{+\infty}\frac{\sin{x}}{x\ln{x}}\, \rm dx$ is conditionally convergent The integral $\displaystyle\int_{2}^{+\infty}\frac{\sin{x}}{x\ln{x}}\,dx$ is conditionally convergent.
I know that $\displaystyle\int_{2}^{+\infty}\frac{\sin{x}}{x}\, dx$
is conditionally convergent and $ {\forall}p > 1$, 
$\displaystyle\int_{2}^{+\infty}\frac{\sin{x}}{x^p}\, dx$ is absolute convergent,
but $\ln{x}$ is between $x$ and $x^p$, so how to prove that $\displaystyle\int_{2}^{+\infty}\frac{\sin{x}}{x\ln{x}}\, dx$ is conditionally convergent?
 A: This results from Abel's test for improper integrals:

If $\;\smash{\displaystyle\int_a^\beta}\!f(x)\,\mathrm d x$ is uniformly bounded over all intervals $[a,\beta]\subset[a,b]$, and $g$ is a  function  decreasing to $0$ on $[a,b]$, then $\;\displaystyle\int_a^{b\strut}\!\! f(x)g(x)\,\mathrm dx\;$ is a convergent improper integral. 

Just take $\;g(x)=\dfrac 1{\ln x}$ and $\;f(x)=\dfrac{\sin x}x$, since it is known that Dirichlet's integral
$$\int_0^{+\infty}\frac{\sin x}x\,\mathrm dx=\frac\pi 2.$$
A: integrate by parts as Lord Shark said.
Since Wu does not know this method, I put it here for future readers of the question.  
Integration by parts shows: for $M > 2$,
$$
\int_2^M\frac{\sin x}{x \log x} dx = \frac{\cos 2}{2 \log 2} - \frac{\cos M}{M \log M}
-\int_2^M \frac{1+\log x}{(x\log x)^2}\;\cos x\;dx
$$
Now
$$
\left|\frac{\cos M}{M\log M}\right| \le \frac{1}{M}\qquad\text{so}\qquad
\lim_{M\to \infty}\frac{\cos M}{M\log M} = 0.
$$
Next,
$$
\left|\frac{1+\log x}{(x\log x)^2}\;\cos x\right| \le \frac{2}{x^2}
\qquad\text{and}\qquad \int_2^\infty\frac{1}{x^2}\;dx\text{ converges}
$$
so
$$
\int_2^\infty \frac{1+\log x}{(x\log x)^2}\;\cos x\;dx\quad\text{converges}
$$
Combining them, we get
$$
\int_2^\infty\frac{\sin x}{x \log x} dx\qquad\text{converges}
$$
A: This is roughly integral analogue of the alternating series test. Since proving its generalization cause little harm, let me actually show

Proposition 1. Suppose that $f : [a, \infty) \to \mathbb{R}$ satisfies the following two conditions:
  
  
*
  
*$f$ is monotone-decreasing, i.e., $f(x) \geq f(y)$ for all $a \leq x \leq y$.
  
*$\lim_{x\to\infty} f(x) = 0$.
  
  
  Then
$$ \int_{a}^{\infty} f(x)\sin(x) \, \mathrm{d}x = \lim_{b\to\infty} \int_{a}^{b} f(x)\sin(x) \, \mathrm{d}x $$
converges. Moreover, this integral is absolutely convergent if and only if $\int_{a}^{\infty} f(x) \, \mathrm{d}x < \infty$.

The proof is quite simple. We first prove that the integral converges. Let $n$ be an integer so that $\pi n \geq a$. Then for $ b \geq \pi n$,
\begin{align*}
\int_{a}^{b} f(x)\sin(x) \, \mathrm{d}x
&= \int_{a}^{\pi n} f(x)\sin(x) \, \mathrm{d}x + \sum_{k=n}^{\lfloor b/\pi\rfloor - 1} \int_{\pi k}^{\pi(k+1)} f(x)\sin(x) \, \mathrm{d}x \\
&\quad + \int_{\pi\lfloor b/\pi\rfloor}^{b} f(x)\sin(x) \, \mathrm{d}x.
\end{align*}
Writing $N = \lfloor b/\pi \rfloor$ and defining $a_k$ by $a_k = \int_{0}^{\pi} f(x+\pi k)\sin(x) \, \mathrm{d}x$, we find that


*

*$a_k \geq 0$, since $f(x+\pi k) \geq 0$ for all $x \in [0, \pi]$.

*$a_{k+1} \geq a_k$ since $f(x+\pi k) \geq f(x+\pi(k+1))$ for all $x \in [0, \pi]$.

*$a_k \to 0$ as $k\to\infty$, since $a_k \leq \int_{0}^{\pi} f(\pi k) \sin (x) \, \mathrm{d}x = 2f(\pi k) \to 0$ as $k \to \infty$.

*Bu a similar computation as in step 3, we check that $\left| \int_{\pi N}^{b} f(x) \sin (x) \, \mathrm{d}x \right| \leq 2f(\pi N)$, and so, $\int_{\pi N}^{b} f(x) \sin (x) \, \mathrm{d}x \to 0$ as $b\to\infty$.

*We have
$$ \sum_{k=n}^{N - 1} \int_{\pi k}^{\pi(k+1)} f(x)\sin(x) \, \mathrm{d}x = \sum_{k=n}^{N-1} (-1)^k a_k. $$
So, by the alternating series test, this converges as $N\to\infty$, hence as $b \to \infty$.
Combining altogether, it follows that $\int_{a}^{b} f(x)\sin(x) \, \mathrm{d}x $ converges as $b\to\infty$.
To show the second assertion, let $n$ still be an integer with $\pi n \geq a$. Then for $k \geq n$, integrating each side of the inequality $f(\pi(k+1))|\sin x| \leq f(x)|\sin x| \leq f(\pi k)|\sin x|$ for $x \in [\pi k, \pi(k+1)]$ gives
$$ 2f(\pi(k+1))
\leq \int_{\pi k}^{\pi(k+1)} f(x)|\sin(x)| \, \mathrm{d}x
\leq 2f(\pi k) $$
and similar argument shows
$$ \pi f(\pi(k+1))
\leq \int_{\pi k}^{\pi(k+1)} f(x) \, \mathrm{d}x
\leq \pi f(\pi k). $$
From this, we easily check that
$$ \frac{2}{\pi} \int_{\pi(n+1)}^{\infty} f(x) \, \mathrm{d}x
\leq \int_{\pi n}^{\infty} f(x)|\sin x| \, \mathrm{d}x
\leq 2f(\pi n) + \frac{2}{\pi} \int_{\pi(n+1)}^{\infty} f(x) \, \mathrm{d}x. $$
Therefore the second assertion follows.
