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Need to solve this integral: $$I=\int_{-1}^{1}dx(\lim_{\varepsilon\to 0^+}\frac{\varepsilon}{\varepsilon^2+x^2}f(x)+\pi\vartheta(x)\frac{df(x)}{dx}(x)) $$ I think I should recognize the limit as a distribution, but I can't find which one with a precise argument, I think it is $ \delta$ Dirac function. So if it is correct this is my solution: $$I=\int_{-1}^{1}\delta(x)f(x)dx+\pi\int_{-1}^{1}\vartheta(x)\frac{df(x)}{dx}(x)dx=$$ $$f(0)+\pi(f(1)-f(0))$$

Is this correct? If the identification of the limit with delta function is correct, can you please tell me why?

Thanks a lot

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  • $\begingroup$ $$\lim_{\varepsilon\to 0^+}\frac{\varepsilon}{\varepsilon^2+x^2} = \pi \delta(x)$$ $\endgroup$ – md2perpe Dec 31 '18 at 13:54
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Almost. The limit is $\pi \delta(x)$ since $$ \int_{-\infty}^{\infty} \frac{\epsilon}{\epsilon^2+x^2} \phi(x) \, dx = \{ x = \epsilon y \} = \int_{-\infty}^{\infty} \frac{\epsilon}{\epsilon^2+\epsilon^2 y^2} \phi(\epsilon y) \, \epsilon \, dy = \int_{-\infty}^{\infty} \frac{1}{1+y^2} \phi(\epsilon y) \, dy \\ \to \int_{-\infty}^{\infty} \frac{1}{1+y^2} \phi(0) \, dy = \left( \int_{-\infty}^{\infty} \frac{1}{1+y^2} \, dy \right) \phi(0) = \pi \phi(0) = \int_{-\infty}^{\infty} \pi \delta(x) \, \phi(x) \, dx $$ for all $\phi \in C_c^\infty.$

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    $\begingroup$ Has the OP observed that the graphical representation of function $f_{\varepsilon}$ defined by $f_{\varepsilon}(x):=\varepsilon/(\varepsilon^2+x^2)$ is obtained from the curve of $f(x):=1/(1+x^2)$ (Cauchy function) by squeezing it in a ratio $1:\varepsilon$ along the $x$-axis while elongating it in an inverse ratio along the $y$ axis, thus preserving the area under its curve. $\endgroup$ – Jean Marie Dec 31 '18 at 19:02

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