What's the deal with this $\frac1\pi$? I recently learned about the very interesting Dirac Delta function, defined as
$$\delta(x)=\frac1\pi\lim_{\epsilon\to 0}\frac{\epsilon}{x^2+\epsilon^2}$$
Which is a very majestic definition, as the function is $0$ everywhere, except for the point $x=0$, at which it is $\infty$. Which brings me to my questions: if this function is $0$ (basically) everywhere, why is there that $\frac1\pi$, and where did it come from?
 A: It's there for normalization, to get
$$\int_{-\infty}^{\infty}\delta(x)\mathrm{d}x=1$$
Just like in the case of approximation with normal distribution:
$$\delta_a(x)=\frac{1}{\sqrt{\pi}a}\exp\left(-\frac{x^2}{a^2}\right)$$
A: This definition doesn't hold in exactly the way that you've written it. It means:
$$f(0)=\lim_{\epsilon \to 0^+} \int_{-\infty}^\infty f(x) \frac{1}{\pi} \frac{\epsilon}{x^2+\epsilon^2} dx.$$
for all sufficiently nice functions $f$. Note that the limit is taken outside the integral; taking the limit inside the integral results in nonsense. Anyway, without the division by $\pi$, you would get $\pi f(0)$, as you can see by taking $f(x)=1$.
We then identify $f(0)$ with the symbol $\int_{-\infty}^\infty f(x) \delta(x) dx$ as a definition of the latter symbol.
A: To expand on Botond's answer, any definition of $\delta(x)$ in the form $\lim_{\epsilon\to 0^+}\frac{1}{\epsilon}\eta\left(\frac{x}{\epsilon}\right)$ with $\eta$ a nascent delta function has $\int_{\Bbb R}\eta(x)dx=1$ so $$\lim_{\epsilon\to 0^+}\int_{\Bbb R}\frac{1}{\epsilon}\eta\left(\frac{x}{\epsilon}\right)dx=\lim_{\epsilon\to 0^+}\int_{\Bbb R}\eta(y)dy=1.$$The example at hand takes $\eta(y)=\frac{1}{\pi}\frac{1}{1+y^2}$, which can be shown with $y=\tan t$ to have the right normalisation.
