How to show $\lim_{\epsilon \rightarrow 0} \frac{\epsilon}{x^2+\epsilon ^2} = \pi \delta (x)$?

Question

Show that :

$\lim_{\epsilon \rightarrow 0} \frac{\epsilon}{x^2+\epsilon ^2} = \pi \delta (x)$

Where $\delta (x)$ is the dirac-delta function.

I can't show that the integral of this over all $x$ is $\pi$

Answer 1

For $x\ne 0$, $$\lim_{\epsilon\to 0}\frac{\epsilon}{x^2+\epsilon^2}=0$$ For $x=0$, $$\lim_{\epsilon\to 0}\frac{\epsilon}{x^2+\epsilon^2}=\infty$$ Besides, $$\int_a^b \frac{\epsilon}{x^2+\epsilon^2}dx=\int \frac{1/\epsilon}{1+(x^2/\epsilon^2)}dx$$ $$=\int_a^b \frac{1}{1+(x^2/\epsilon^2)}d(x/\epsilon)$$ $$=\arctan(x)\bigg|_{a/\epsilon}^{b/\epsilon}$$ Now if $a>0, b>0$, then $$\lim_{\epsilon\to infty}\arctan(x)\bigg|_{a/\epsilon}^{b/\epsilon}=\pi/2 - \pi/2 =0$$ Similary, for $a<0, b<0$, we have $$\lim_{\epsilon\to infty}\arctan(x)\bigg|_{a/\epsilon}^{b/\epsilon}=(-\pi/2) - (-\pi/2) =0$$ For $a<0, b>0$, we have $$\lim_{\epsilon\to infty}\arctan(x)\bigg|_{a/\epsilon}^{b/\epsilon}=\pi/2 - (-\pi/2) =\pi$$ Hence $$\lim_{\epsilon\to 0}\frac{\epsilon}{x^2+\epsilon^2}=\pi\delta(x)$$

Answer 2

Since you tagged distribution theory, I think the following should work.

For any test function $\varphi(x)\in\mathcal{D}(\mathbb{R})$, where $\mathcal{D}(\mathbb{R})$ consists of $C^{\infty}$-smooth functions with compact support (which indicates that $\lim_{x\to\pm\infty}\varphi(x)=0$). We have the following $$ \begin{aligned} \int_{-\infty}^{\infty}\frac{\varepsilon\varphi(x)}{x^{2}+\varepsilon^{2}}dx&=\int_{-\infty}^{\infty}\varphi(x) d\tan^{-1} \left(\frac{x}{\varepsilon}\right) \\ &= \left[ \varphi(x)\tan^{-1} \left(\frac{x}{\varepsilon}\right) \right]_{-\infty}^{\infty}-\int_{-\infty}^{\infty}\tan^{-1} \left(\frac{x}{\varepsilon}\right)d\varphi \end{aligned} $$ then we have $$ \begin{aligned} \lim_{\varepsilon\to0}\int_{-\infty}^{\infty}\frac{\varepsilon\varphi(x)}{x^{2}+\varepsilon^{2}}dx&= -\lim_{\varepsilon\to0}\int_{-\infty}^{\infty}\tan^{-1} \left(\frac{x}{\varepsilon}\right)d\varphi \\ &=-\frac{\pi}{2}\int_{0}^{\infty}d\varphi+\frac{\pi}{2}\int_{-\infty}^{0}d\varphi \\ &=-\frac{\pi}{2} \Big[ \varphi(x) \Big]_{0}^{\infty}+ \frac{\pi}{2} \Big[ \varphi(x) \Big]_{-\infty}^{0} \\ &=\pi\varphi(0) \end{aligned} $$ then $$\lim_{\varepsilon\to0}\frac{\varepsilon}{x^{2}+\varepsilon^{2}}=\pi\delta(x)$$ in the distributional sense.

Answer 3

Note that in the classical sense

$$\lim_{\epsilon \to 0}\frac{\epsilon}{x^2+\epsilon^2}=\begin{cases}0&,x\ne 0\\\\\text{undefined}&,x=0\end{cases}$$

In the sense of generalized functions, the expression

$$\lim_{\epsilon \to 0}\frac{\epsilon}{x^2+\epsilon^2}\sim \pi \delta(x)$$

means that for all suitable test functions $\phi$ we have

$$\lim_{\epsilon \to 0}\int_{-\infty}^\infty \left(\frac{\epsilon}{x^2+\epsilon^2}\right)\,\phi( x)\,dx=\pi \phi(0) \tag 1$$

To show that $(1)$ is correct we let $\phi$ be a suitable test function. Then, we can write

$$\begin{align} \int_{-\infty}^\infty \left(\frac{\epsilon}{x^2+\epsilon^2}\right)\,\phi(x)\,dx&=\phi(0)\int_{-\infty}^\infty \left(\frac{\epsilon}{x^2+\epsilon^2}\right)\,dx+\int_{-\infty}^\infty \left(\frac{\epsilon}{x^2+\epsilon^2}\right)\,\left(\phi(x)-\phi(0)\right)\,dx\\\\ &=\pi\,\phi(0)+\int_{-\infty}^\infty \left(\frac{1}{x^2+1}\right)\,\left(\phi(\epsilon x)-\phi(0)\right)\,dx \end{align}$$

The Dominated Convergence Theorem guarantees that

$$\begin{align} \lim_{\epsilon \to 0}\int_{-\infty}^\infty \left(\frac{1}{x^2+1}\right)\,\left(\phi(\epsilon x)-\phi(0)\right)\,dx&=\int_{-\infty}^\infty \left(\frac{1}{x^2+1}\right)\,\lim_{\epsilon \to 0} \left(\phi(\epsilon x)-\phi(0)\right)\,dx\\\\ &=0 \end{align}$$

Therefore, we find that for all test functions $\phi(x)$

$$\lim_{\epsilon \to 0}\int_{-\infty}^\infty \left(\frac{\epsilon}{x^2+\epsilon^2}\right)\,\phi( x)\,dx=\pi \phi(0)$$

as was to be shown!