limit of an integral with a Lorentzian function We want to calculate the $\lim_{\epsilon \to 0} \int_{-\infty}^{\infty} \frac{f(x)}{x^2 + \epsilon^2} dx $ for a function $f(x)$  such that $f(0)=0$. We are physicist, so the function $f(x)$ is smooth enough!.
After severals trials, we have not been able to calculate it except numerically. 
It looks like the normal Lorentzian which tends to the dirac function, but a 
$\epsilon$ is missing.
We wonder if this integral can be written in a simple form as function of $f(0)$ or its derivatives $f^{(n)}(0)$ in 0. 
Thank you very much. 

I'll assume that $f$ has compact support (though it's enough to suppose that $f$ decreases very fast). As $f(0)=0$ he have $f(x)=xg(x)$ for some smooth $g$. Let $g=h+k$, where $h$ is even and $k$ is odd. As $k(0)=0$, again $k(x)=xm(x)$ for some smooth  $m$.
We have $$\int_{-\infty}^{\infty} \frac{f(x)}{x^2 + \epsilon^2} dx =\int_{-\infty}^{\infty} \frac{xg(x)}{x^2 + \epsilon^2} dx =\int_{-\infty}^{\infty} \frac{x(h(x)+xm(x))}{x^2 + \epsilon^2} dx = \int_{-\infty}^{\infty} \frac{x^2m(x)}{x^2 + \epsilon^2} dx $$
(the integral involing $h$ is $0$ for parity reasons)
and
$$\int_{-\infty}^{\infty} \frac{x^2m(x)}{x^2 + \epsilon^2} dx=\int_{-\infty}^{\infty} m(x)dx-\int_{-\infty}^{\infty} \frac{m(x)}{(x/\epsilon)^2 + 1}  dx. $$
The last integral converges to $0$, so the limit is
$\int_{-\infty}^{\infty} m(x)dx$
where (I recall)
$$m(x)=\frac{f(x)+f(-x)}{2x^2}.$$
