I find integrals on the real axes computed by complex number techniques and looking for a generalization working just on the positive semiaxes. I tried to put together this general result, but I am not sure if I made mistakes or if my reasonings are correct, maybe with some more assumptions.
CLAIM: Let $f(x):\Bbb{R}^+\rightarrow \Bbb{R}$ be a real function with meromorphic extension $f(z)$ analytic on the real positive axes and going to zero faster than $1/|z|^2$. Than:
$\int_0^{+\infty} f(x) dx = - \lim_{a\rightarrow 0^+} \frac{1}{a} \sum Res_{\Bbb{C} \setminus \Bbb{R}^+} (f(z)z^a)$
where we exclude possible residues at infinity, on the real axes, and $z^a$ refers to the principal branch, real on the real positive axes.
Proof: First we write:
$\int_0^{+\infty} f(x) dx=\lim_{a\rightarrow 0^+} \int_0^{+\infty} f(x)x^a dx [1]$.
For small $a$, if $f$ goes to zero faster than $x^2$, hopefully this operation should be correct.
We can now take as a contour in the complex plane $C=C_1+C_2+C_3$, where $C_1=[0,+R]$, $C_2=Re^{[0,2\pi i]}$, $C_3=[+R,0]$, where $C_1$ is slightly above the real axes and $C_3$ slightly below. Note that $f(z)z^a$ is not analytic and we use the usual branch cut of the logarithm to define it. $z^a$ in our choice is real therefore on $C_2$.
We can now apply the residue theorem in a standard way, letting $R$ go to infinity:
$2\pi i\sum Res_{\Bbb{C} \setminus \Bbb{R}^+}(f(z)z^a)=\int_C f(z)z^a= (1-e^{i2\pi a})\int_0^{+\infty} f(x)x^a dx$
dividing by the prefactor, using [1] and expanding the exponential we find the thesis.
Anyone can tell me if the result/proof is correct or suggest improvements? Maybe the claim is true under some more assumptions ? Please have some patience if I make mistakes but I am trying to get more used to complex analysis...