Considering the two following integrals $$n\in \mathbb{Z}$$ $$\int_{-\infty}^{\infty}\frac{x^{n}}{(x+2i)(x-2i)}$$
$$\int_{-\infty}^{\infty}\frac{x^{n}e^{i\pi x}}{(x+2i)(x-2i)}$$
I do not know what conditions to impose so that they converge. I know that they do not have the same result but do not know what's the difference in this case is to add the term $e^{i\pi x}$.
Thank you so much for your help.
In οrder to calculate this integral first you need to take into account :$$\int_{\Gamma_{R}} \frac{z^n}{(z+2i)(z-2i)}dz= \int_{-R}^{R}\frac{x^n}{x^2+4}dx+\int_{C_R}\frac{z^n}{z^2+4}dz$$ Where $\Gamma_{R}$ is a semicircle of radius $R$ and diameter the straight $[-R,R]$, and $C_{R}$ is the semicircle contour without the straight line. We need to see that our second integral $$\int_{C_R}f(z)dz\to0$$ In order to prove this one has to use the estimation lemma (a.k.a M-L inequality).
$$\left|\int_{C_R}\frac{z^n}{z^2+4}dz\right| \le \max_{|z|=R}\left|\frac{z^n}{z^2+4}\right|\pi R=\frac{\pi R R^n}{R^2+4} \le \frac{R^{n+1}}{R^2}$$.Take $R\to \infty$ and then your integral tends to zero whenever $n+1<2$.Then you can use the residue theorem and you are done!