# 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?

• integrate by parts? – Lord Shark the Unknown Apr 7 at 11:27
• I don't know how to do that... I have tried to put $\frac{1}{x}$ into $dx$, but $\left| \sin{x} \right|$ is unable to integrate. How to do that? – Wu Matt Apr 7 at 11:36

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.

• Thanks for your answer!!! – Wu Matt Apr 7 at 12:07

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

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