Limit of Lorentzian is Dirac Delta I have a quick question that just came up in my research and I could not find an answer anywhere so I thought I'd try here.
So one of the definitions of the Dirac Delta is the limit of the Lorentzian function with $\epsilon$ going to zero. See here http://hitoshi.berkeley.edu/221a/delta.pdf for the expression on the first page.
My question is, can I define the Dirac Delta just as well with this
$$\delta(t) ~=~ \lim_{\epsilon\rightarrow 0} \frac{1}{\pi}\frac{\epsilon^2}{\epsilon^2+t^2},$$ 
where I have included an extra $\epsilon$ in the numerator. My hunch is that this is no problem since the limiting behavior looks the same to me. 
 A: It looks like a delta-function. However, because $\epsilon / (\epsilon^2+t^2)$ - you should omit one $\epsilon$ in the numerator, to get the right integral equal to one, by the way - decreases too slowly as $|t|\to\infty$, as $1/t^2$, it will only work as the Dirac delta distribution for test functions that decrease at infinity or at least increase slower than as $O(t)$. If the test function is $t^2$, for example, the integral
$$ \int_{-\infty}^\infty dt\,t^2\,\delta(t) $$
should yield 0 because $t^2=0$ for $t=0$. However, with your definition of the delta function, you will get a divergent answer because the infinite-range integral ultimately beats any $\epsilon$. For this reason, one usually wants approximations of delta functions that decrease faster at $|t|\to\infty$ than the Lorentzian.
Obviously, if you include one extra $\epsilon$, you get $\epsilon\cdot \delta(t) = 0$ regardless of details about the $|t|\to\infty$ behavior.
A: The answer is no, the generalized function (=distribution) 
$$ \lim_{\epsilon\rightarrow 0} \frac{\epsilon^2}{\epsilon^2+t^2}=0 
\qquad\mathrm{a.e.}
$$ 
is almost everywhere (a.e.) equal to the ordinary zero function $0:\mathbb{R}\to\mathbb{R}$ that sends $t\mapsto 0$.
Proof. Consider a test function $f\in C^{\infty}_c(\mathbb{R})$, i.e., an infinitely often differentiable function $f$ with compact support. Then 
$$\int_{\mathbb{R}} dt \ f(t)\frac{\epsilon^2}{\epsilon^2+t^2}  ~\stackrel{t=\epsilon x}{=}~\epsilon \cdot \int_{\mathbb{R}} dx \ f(\epsilon x)\cdot\frac{1}{1+x^2} $$
$$ \longrightarrow 0\cdot f(0)\cdot\int_{\mathbb{R}} dx \ \frac{1}{1+x^2} = 0 \cdot f(0)\cdot\pi =0 \qquad \mathrm{for} \qquad \epsilon \to 0,$$
because of, e.g., Lebesgue's dominated convergence theorem. $\Box$ 
The distribution only becomes $\pi\delta(t)$, if we remove one factor of $\epsilon$. Here $\delta(t)$ is the Dirac delta distribution (often called the Dirac delta function).
Instead of using distribution theory, we may simply interpret the formula
$$ \lim_{\epsilon\rightarrow 0} \frac{\epsilon^2}{\epsilon^2+t^2}~=~\delta_{t,0}~=~\left\{\begin{array}{rcl} 1 &\mathrm{for}& t=0 \\  0 &\mathrm{for}& t\in\mathbb{R}\backslash\{0\} \end{array}\right. $$ 
as a $t$-pointwise limit. Here $\delta_{t,0}$ is the Kronecker delta function, which should not be confused with the Dirac delta distribution. The former is an ordinary function, while the latter is not. The last formula has the added benefit, that it is true both in a $t$-pointwise sense and in a distribution sense, since the Kronecker delta function is zero almost everywhere (with respect to the Lebesgue measure).
A: Let us first show the following where $\delta$ denotes the Dirac delta distribution as usual. If $\{L_\gamma\}_{\gamma > 0}$ is an integrable family of functions with the properties

*

*$\displaystyle \int_{-\infty}^\infty L_\gamma(x) \, dx = 1$ for all $\gamma > 0$ (i.e., $L_\gamma$ are probability distributions),

*$\displaystyle \lim_{\gamma \to 0} \int_{-\epsilon}^\epsilon L_\gamma(x) \, dx = 1$ for all $\epsilon > 0$,

then $L_\gamma \to \delta$ as $\gamma \to 0$, as distributions.
Mathematically, this translates to the goal of showing that for all test functions $f \in C_\mathrm{c}^\infty(\mathbb R)$, we have
$$
\lim_{\gamma \to 0} \int_{-\infty}^\infty L_\gamma(x)f(x) \, dx = f(0)
$$
given the two properties above.
Proof:
Start by taking any $\epsilon > 0$. Then
\begin{align*}
\lim_{\gamma \to 0} \int_{-\infty}^\infty L_\gamma(x)f(x) \, dx &= \lim_{\gamma \to 0} \int_{-\epsilon}^\epsilon L_\gamma(x)f(x) \, dx + \lim_{\gamma \to 0} \int_{-\infty}^{-\epsilon} L_\gamma(x)f(x) \, dx + \lim_{\gamma \to 0} \int_{\epsilon}^\infty L_\gamma(x)f(x) \, dx \\
&= \lim_{\gamma \to 0} \int_{-\epsilon}^\epsilon L_\gamma(x)f(x) \, dx + O\left(\lim_{\gamma \to 0} \int_{-\infty}^{-\epsilon} L_\gamma(x) \, dx + \lim_{\gamma \to 0} \int_{\epsilon}^\infty L_\gamma(x) \, dx\right) \\
&= \lim_{\gamma \to 0} \int_{-\epsilon}^\epsilon L_\gamma(x)f(x) \, dx.
\end{align*}
The second equality uses boundedness of $f$ and the third equality uses properties 1 and 2. Since $\epsilon > 0$ was arbitrary, taking limits give
\begin{align*}
\lim_{\gamma \to 0} \int_{-\infty}^\infty L_\gamma(x)f(x) \, dx 
&= \lim_{\epsilon \to 0}\lim_{\gamma \to 0} \int_{-\epsilon}^\epsilon L_\gamma(x)f(x) \, dx \\
&= \lim_{\epsilon \to 0}\lim_{\gamma \to 0} \int_{-\epsilon}^\epsilon L_\gamma(x)f(0) \, dx + \lim_{\epsilon \to 0}\lim_{\gamma \to 0} o(\epsilon) \\
&= f(0) \cdot \lim_{\epsilon \to 0}\lim_{\gamma \to 0} \int_{-\epsilon}^\epsilon L_\gamma(x) \, dx \\
&= f(0)
\end{align*}
as desired. The second equality uses continuity of $f$ and the last equality uses property 2.
Now, let us apply the above for the Lorentzian function (also called Cauchy/Lorentz/Cauchy-Lorentz distribution):
$$
L_\gamma(x) = \frac{1}{\pi}\left(\frac{\gamma}{\gamma^2 + x^2}\right).
$$
Property 1 can be verified by a straightforward integration using the antiderivative in terms of $\arctan$. Similarly, the explicit antiderivative can be used to verify property 2. Often times, explicit integration may not be possible in order to verify property 2, so the following alternative argument is helpful. Note that $L_\gamma|_{\mathbb R \setminus \{0\}} \to 0$ as $\gamma \to 0$. For all $x > 0$, taking the derivative of $\gamma \mapsto L_\gamma(x)$, we find that its maximum is at $\gamma = x$ and so this map is increasing on $(0, x]$. Thus, $\{L_\gamma|_{[\epsilon, \infty)}\}_{\gamma \in (0, \epsilon]}$ is a monotone increasing family of functions. Applying the dominated convergence theorem gives $\lim_{\gamma \to 0} \int_\epsilon^\infty L_\gamma(x) \, dx = 0$. By symmetry, we also get $\lim_{\gamma \to 0} \int_{-\infty}^{-\epsilon} L_\gamma(x) \, dx = 0$. Finally, this together with property 1 implies the desired property 2.
