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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.

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    $\begingroup$ Any construction where you have some $f(x,\epsilon)$ where $\int_{-\infty}^{+\infty}f\,dx =1$ for all $\epsilon$ and where $\lim_{\epsilon\rightarrow 0}f$ is zero for all $x\neq 0$ will give you the Dirac delta function. $\endgroup$ May 8, 2011 at 5:50
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    $\begingroup$ @JerrySchirmer This is not true. For example take a compactly supported bump function and suppose the bump travels and "carries the mass" to infinity w.r.t. your $\epsilon$ parameter. Then its limit w.r.t. $\epsilon$ gives the zero distribution. $\endgroup$ Jul 6, 2022 at 21:56

3 Answers 3

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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).

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    $\begingroup$ The OP should also compare this again Jack's comment (as they are isomorphic in spirit). Using the change of variables formula you can show that as $\epsilon\to 0$ the "total integral" of $\epsilon^2/(\epsilon^2 + t^2)$ decreases to 0. It comes from exactly the "pulling" out of an $\epsilon$ as in the first equality in the Qmechanic's displayed formula. $\endgroup$ May 8, 2011 at 19:58
  • $\begingroup$ thanks for the answer. So since the expression in question is NOT a form for the Delta function, it will be zero even if t=0. Is that right? $\endgroup$
    – BeauGeste
    May 8, 2011 at 23:28
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    $\begingroup$ @BeauGeste: $\int_{-t_{0}}^{t_{0}}\,dt \frac{e^{2}}{e^{2}+t^{2}}$ comes out to $2e\tan^{-1}(\frac{t_{0}}{e})$. If you take the limit of $t_{0}$ to infinity first, this works out to just be $2\,e\pi$, which goes to zero as $e\rightarrow 0$. If you take the limit of $e\rightarrow 0$ first, you find that the limit is zero for all $t_{0} \neq 0$. $\endgroup$ May 9, 2011 at 4:37
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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.

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  • $\begingroup$ thanks for the response. If I understand you correctly, you are also saying that the Lorentzian is a poor choice to define the Delta function by? Anyways, my test function is unity so I suppose I am fine considering 'my' definition to be ok for the Delta function. $\endgroup$
    – BeauGeste
    May 8, 2011 at 5:20
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    $\begingroup$ I have no idea what you are talking about. $t^2$ is not a test function by any means. A test function is either a smooth function with compact support (if you consider all distributions) or a Schwartz class function (if you consider tempered distributions). $\endgroup$ May 8, 2011 at 19:55
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    $\begingroup$ Maybe in maths, Willie. But physicists integrate $\delta$ functions with any other factors and they almost never encounter functions whose support is strictly compact. This is a physics forum so the relevance of your comment is strictly equal to zero. $\endgroup$ May 9, 2011 at 4:52
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    $\begingroup$ @LubošMotl There is no need to be rude. WillieWong's comment, although technical, is very relevant. If one uses math and it's wrong, then it's wrong. Although physicists often write nonrigorous expressions, they stem from good intuition and can actually be made mathematically rigorous (and if it could not be, then something will fail in the physics as well). May be this post is helpful: math.stackexchange.com/questions/1257447/…. $\endgroup$ Jul 6, 2022 at 22:04
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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

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

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