Show that $\int\limits_2^{+\infty}\frac{\sin{x}}{x\ln{x}}\, \rm dx$ is conditionally convergent

Question

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 1

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 $$\int_0^{+\infty}\frac{\sin x}x\,\mathrm dx=\frac\pi 2.$$

Answer 2

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 $$ \int_2^\infty\frac{\sin x}{x \log x} dx\qquad\text{converges} $$

Answer 3

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.