I'm reading a paper (in physics, relativistic QFT) that makes the following assertion: \begin{align} \int_{-\infty}^\infty \mathrm{d} s \left( \frac{1}{s^2} \right) = 0 \end{align} It is simply stated that it vanishes by contour integration. Is this supposed to be obvious? Intuitively, I don't see how this is the case since it is an even function on an even domain. Does anyone know the solution? The statement can be found after Equation (5.7) here: https://arxiv.org/abs/gr-qc/0606067
By extension could one then argue that: \begin{align} \int_{-\infty}^\infty \mathrm{d} s \left( \frac{f(s)}{s^2} \right) = 0 \end{align} where $f(s)$ is even in $s$?
As stated, this is wrong. The function $\frac{1}{s^2}$ has a pole at $0$ and is not integrable there. What they probably mean (hinted at with the word contour integration) is the claim $$ \lim_{\varepsilon \rightarrow 0} \int_{-\infty +i\varepsilon}^{\infty +i\varepsilon} \frac{1}{(s+i\varepsilon)^2}ds = \lim_{\varepsilon \rightarrow 0} \lim_{s \rightarrow \infty} (-\frac{1}{s + i \varepsilon} + \frac{1}{-s + i \varepsilon})=\lim_{\varepsilon \rightarrow 0} \lim_{s \rightarrow \infty}\frac{2s}{-s^2-\varepsilon^2}=0$$