# An exemple of integral of distributions

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

• $$\lim_{\varepsilon\to 0^+}\frac{\varepsilon}{\varepsilon^2+x^2} = \pi \delta(x)$$ – md2perpe Dec 31 '18 at 13:54

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 $$\phi \in C_c^\infty.$$
• 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. – Jean Marie Dec 31 '18 at 19:02