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. 
 A: 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)
$$m(x)=\frac{f(x)+f(-x)}{2x^2}.$$
A: Let $f$ be smooth with compact support. Consider the double layer potential (up to a constant)
$$
u(x_1,x_2)=-2\pi\int_{-\infty}^{\infty}\frac{\partial}{\partial x_2}\Gamma(x_1-y_1,x_2)f(y_1)\,dy_1=
$$ 
$$
\int_{-\infty}^{\infty} \frac{x_2f(y_1)}{x^2 + x_2^2} dx,
$$
where $\Gamma(x)=-\frac1{2\pi}\log|x|$ is a fundamental solution for the Laplace equation.
As is known $u$ is smooth  up to the boundary for smooth $f$. We have
$$
\frac{\partial u(0,0)}{\partial x_2}=
\lim_{\epsilon \to 0+} \frac{u(0,\epsilon)-u(0,0)}\epsilon= 
\lim_{\epsilon \to 0+} \frac{u(0,\epsilon)-f(0)}\epsilon= 
\lim_{\epsilon \to 0+}\int_{-\infty}^{\infty} \frac{ f(y_1)}{x^2 + \epsilon^2} dx,
$$
which is the required value. 
To calculate it note that 
$$
\frac{\partial u(x_1,x_2)}{\partial x_2} =
-2\pi\int_{-\infty}^{\infty}\frac{\partial^2}{\partial x_2^2}\Gamma(x_1-y_1,x_2)f(y_1)\,dy_1=
$$
$$
2\pi\int_{-\infty}^{\infty}\frac{\partial^2}{\partial y_1^2}\Gamma(x_1-y_1,x_2)f(y_1)\,dy_1=
2\pi\int_{-\infty}^{\infty}\Gamma(x_1-y_1,x_2)f''(y_1)\,dy_1
$$
because $\Gamma$ satisfies the Laplace equation.
The last integral converges uniformly for $|x|\le 1$ so taking the limit $x\to0$ gives
$$
\frac{\partial u(0,0)}{\partial x_2}=2\pi\int_{-\infty}^{\infty}\Gamma(0-y,0)f''(y)\,dy_1=-\int_{-\infty}^{\infty}\log|y|f''(y)\,dy.
$$
