I would like to check whether

$$\int_{0}^{\infty} \frac{x \sin(x)}{1+x²}dx$$

converges and converges absolutely.

I have a feeling that neither is true, however none of the methods known to me seem to help. I struggle to find a lower estimate for the function. Any hints and help welcome.

I tried using $$\frac{x \sin(x)}{1+x²}\leq \frac{x \sin(x)}{x²}=\frac{ \sin(x)}{x}$$

of which I know that it absolutely converges, but it only holds for when $\sin(x)\geq0$, so it does not help with the non-absolute convergence.

  • $\begingroup$ For the absolute convergence: do you know whether $\int_0^{\infty}\frac{\lvert\sin x\rvert}{x}\,dx$ converges? $\endgroup$ – Clement C. Mar 26 '17 at 16:07
  • $\begingroup$ Yes, I do, I have added it to the question. $\endgroup$ – B.Swan Mar 26 '17 at 16:08
  • $\begingroup$ I cannot see where you added that. But if you know that part, given that $\frac{\lvert\sin x\rvert}{x}\sim_{x\to\infty}\frac{\lvert x\sin x\rvert}{1+x^2}$, you can conclude. $\endgroup$ – Clement C. Mar 26 '17 at 16:09

Note that for $x\ge 1$, $\frac{x}{1+x^2}$ monotonically decreases to $0$ and $\int_0^L \sin(x)\,dx\le 2$ for all $L$.

Therefore, Abel's (Dirichlet's) Test for improper integrals guarantees that the integral coverges.

To show that we have conditional convergence only, we have $\frac{x}{1+x^2}\ge \frac{1}{2x}$ for $x\ge 1$.

Then, we can write

$$\begin{align} \int_{1}^{(n+1)\pi}\frac{|\sin(x)|}{x}\,dx&\ge\sum_{k=1}^n\int_{k\pi}^{(k+1)\pi}\frac{|\sin(x)|}{x}\,dx\\\\ &\ge \sum_{k=1}^n \frac{1}{(k+1)\pi}\int_0^\pi \sin(x)\,dx\\\\ &=\sum_{k=1}^n \frac{2}{(k+1)\pi}\tag 1 \end{align}$$

Since, the series in $(1)$ diverges by comparison with the harmonic series, then the integral of interest does not absolutely converge.

  • $\begingroup$ You should probably say that $x/(1+x^2)$ is eventually monotonically decreasing, since it's zero when $x=0$. $\endgroup$ – Chappers Mar 26 '17 at 16:13
  • $\begingroup$ @Chappers I've edited. Thank you! And nice to see you're still here. -Mark $\endgroup$ – Mark Viola Mar 26 '17 at 16:15
  • $\begingroup$ Note that the integrand is positive. Thus, for $b>a+1$, $\int_a^b f(x)\,dx\ge \int_{a+1}^bf(x)\,dx$. $\endgroup$ – Mark Viola Mar 26 '17 at 17:38
  • $\begingroup$ I may miss the point. To be precise I ask about how you changed the limits of the integral from $k\pi, (k+1)\pi$ to $0, \pi$. Is it because the integral is identical between those points? $\endgroup$ – B.Swan Mar 26 '17 at 17:44
  • $\begingroup$ I see. $|\sin(x)|$ is $\pi$-periodic. $\endgroup$ – Mark Viola Mar 26 '17 at 17:49

Let $f(x)=(x\sin x)/(1+x^2).$

(1). The sign of $f(x)$ is $(-1)^{n-1}$ for $x\in ((n-1)\pi,n\pi)$, for any $n\in \mathbb N.$

For $n\in \mathbb N$ let $I_n=|\int_{(n-1)\pi}^{n\pi}f(x)\;dx|.$

For $y\geq 0$ let $n_y\pi$ be the largest integer multiple of $\pi$ that does not exceed $y$. We have:

(2). $|\int_{n_y\pi}^yf(x)\;dx|< I_{(1+n_y)}$ for $y\geq 0$, by (1).

(3). $|\int_0^y f(x)\;dx-\sum_{j=1}^{n_y}(-1)^{j-1}I_j|=|\int _{n_y\pi}^y f(x)\;dx|<|I_{(1+n_y)}|$ for $y\geq 0$, by (2).

(4). $I_{n+1}<I_n$ because of (1) and because $x\in ((n-1)\pi,n\pi) \implies |f(x+\pi)|=\frac {|x+\pi|}{1+(x+\pi)^2}|\sin x|<|f(x)|.$

(5). We have $\lim_{n\to \infty} I_n=0$ because for $n\geq 2$ we have $0<I_n<\int_{(n-1)\pi}^{n\pi}(1/x)\;dx<\frac {1}{n-1}.$

(6). $\sum_{j=1}^{\infty}(-1)^jI_j$ converges by (4) and (5).

Applying (5) and (6) to (3) we have $\int_0^{\infty}f(x)\;dx=\sum_{j=1}^{\infty}(-1)^jI_j.$

The non-convergence of $\int_0^{\infty}|f(x)|\;dx$ is covered in another answer.


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