Determining whether $\int_{0}^{\infty} \frac{x \sin(x)}{1+x^2}dx$ converges and converges absolutely 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.
 A: 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.
A: 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.
