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For some nice function $f(x)$, how to show the following using contour integration? $$ \lim_{\epsilon\to 0^+} \int_{-\infty}^\infty f(x) \frac{\epsilon}{\epsilon^2+x^2} dx = \pi f(0) $$

The way I would approach this integral is by $\frac{d}{dx} \tan^{-1} \frac{x}{\epsilon} = \frac{\epsilon}{\epsilon^2+x^2}$ and integrating by parts. This assumes $f$ does not increase too quickly, as for example when $f(x)=x$, the integral on the LHS does not converge.

I'm hoping to see a method that can be extended to an integral of the form $$ \lim_{\epsilon\to 0^+} \int_{-\infty}^\infty f(x) \frac{\epsilon}{(\epsilon^2+x^2)(x^2 - (\omega+i\epsilon)^2)} dx $$

where $f$ has some poles in the complex plane.

Also found a similar question here: Dirac delta as a limit of sequence of functions

Bio
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    I would change the variable $x=\epsilon u$ and with the right niceness assumptions on $f$, apply the Lebesgue Dominated Convergence Theorem. – Stefan Lafon Oct 10 '23 at 20:49
  • Just realized this is the same as the answer that you linked in your post. – Stefan Lafon Oct 10 '23 at 20:50
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    Also, I think there's a sign error. If you replace $f$ by the constant function $1$, you should get $\pi$ in the right-hand side. Or another way to see it, is that if $f$ is positive almost everywhere, so should the right-hand side be. – Stefan Lafon Oct 10 '23 at 20:51
  • Thanks, fixed the typo – Bio Oct 11 '23 at 07:18
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    To do this by contour integration, you'd have to assume $f$ is analytic over either the entire upper or lower half-plane, and that $\frac {f(z)}z \to 0$ as $z \to \infty$. This is far too limited a set of functions for such a proof to be worthwhile. – Paul Sinclair Oct 11 '23 at 17:43

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