$\int_{0}^{+\infty}\frac{\sin x}{x^{k}(1+x^{2})}dx \$ via residue calculus

I want to evaluate with calculus of residues

$$\int_{0}^{+\infty}\frac{\sin x}{x^{k}(1+x^{2})}dx \$$

$k \in \mathbb{N}, k \geq 1$

If $k = 1$ we have $$\int_{0}^{+\infty}\frac{\sin x}{x(1+x^{2})}dx = \frac{1}{2}\int_{-\infty}^{+\infty}\frac{\sin x}{x(1+x^{2})}dx$$ and we can integrate $$f(z) = \frac{e^{iz}}{z(1+z^{2})}$$ along the well known semi-circumference with a "bite" in $0$ and apply Jordan lemma. But what about the case $k > 1$ ?

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1 Answer

The integral does not converge for k>1 assuming k is an integer. Are you interested in fractional values of k? If so you can use intended paths, except now we will have a branch cut.

You can show the integral is not convergent for k>1 by writing

$$\int_0^\infty \frac{\sin x}{x^{k}(x^2+1)}dx=\int_0^1 \frac{\sin x}{x^{k}(x^2+1)}dx + \int_1^\infty \frac{\sin x}{x^{k}(x^2+1)}dx$$ and now treat each integral separately, write $\sin x$ as it's series and you will be able to see that $k>1$ is divergent.

If you are concerned with fractional values of k, let me know, we can work that out.

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