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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.

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We are physicist sounds like ""We are Borg. Resistance is futile..." – draks ... Jul 17 '12 at 12:42
We may have a problem at $0$: for example, if $\chi$ is a bump function (smooth, with compact support, non-negative, and constant equal to $1$ in a neighborhood of $0$) and $f(x)=x\chi(x)$, the limit is infinite. But it's finite if we add the condition $f'(0)=0$, as Taylor's formula shows. – Davide Giraudo Jul 17 '12 at 12:49
Physicists (I am myself one) tend to replace this integral with $ (\pi/\epsilon) f (0) $ (letting $\epsilon$ not going all the way to 0 ;-). However, one needs more than smoothness to justify this. – Fabian Jul 17 '12 at 13:02
Has the function $f$ poles in the complex plane? Is it analytic? For an entire function the result would be $\frac{\pi f(i\epsilon)}{\epsilon}$ (well if the integral on the circular contour in the half plane goes to $0$ of course). – Raymond Manzoni Jul 17 '12 at 13:30

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) $$m(x)=\frac{f(x)+f(-x)}{2x^2}.$$

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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 $$ \frac{\partial u(0,0)}{\partial x_2}=2\pi\int_{-\infty}^{\infty}\Gamma(0-y,0)f''(y)\,dy_1=-\int_{-\infty}^{\infty}\log|y|f''(y)\,dy. $$

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