integration - Show that $intlimits_2^{+infty}frac{sin{x}}{xln{x}}, rm dx$ 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?

Answer

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:

1. $f$ is monotone-decreasing, i.e., $f(x) \geq f(y)$ for all $a \leq x \leq y$.


2. $\lim_{x\to\infty} f(x) = 0$.

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

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





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




3. $a_k \to 0$ 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 $k \to \infty$.




4. 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 $b\to\infty$.




5. 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.