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i have got a question to this integral: How would one evaluate $\int\frac{x\sin(x)}{x^2+1}$ over the real line?

I can calculate it easily with residue theory like there. Now i want to see convergence (or absolute convergence?) without calculating. How can i do that? The integral ist now over half the real line:

$\int_0^\infty \frac{x\sin(x)}{1+x^2}dx$

Is it a good idea to write the integral in the form

$\int_0^\infty \frac{x\sin(x)}{1+x^2}dx=\int_0^1 \frac{x\sin(x)}{1+x^2}dx+\int_1^\infty \frac{x\sin(x)}{1+x^2}dx$ ?

Sorry for "bad" english.

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    Conditional convergence as $x\to\infty$. – Simply Beautiful Art Jan 30 '17 at 12:36
  • Using Abel-Dirichlet criteria to get convergence without absolute value. Then using the following to get divergence under the abs value symbol: $$\dfrac{x}{1+x^2}\sim \frac1x$$ and $$|\sin x|\ge \frac12,\quad\forall x\in[\frac\pi 6+k\pi,\frac{5\pi}6+k\pi].$$ – Vim Jan 30 '17 at 12:44

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