1
$\begingroup$

From Wikipedia I found this about the delta-function

$$\delta(x)=\begin{cases}\infty,\:\:x=0\\0,\:\:x\neq 0\end{cases}$$

$$\int_\limits{-\infty}^{\infty}\delta(x)dx=1$$

I tried to prove that $$\int_\limits{-\infty}^{\infty}\delta(x)dx=1$$.

$\int_\limits{-\infty}^{\infty}\delta(x)dx=\int_\limits{-\infty}^{0}\delta(x)+\int_\limits{0}^{+\infty}\delta(x)$

However, I do not see how these integrals are going to yield $1$ as a result.

Question:

How do I prove $\int_\limits{-\infty}^{\infty}\delta(x)dx=1$?

Thanks in advance!

$\endgroup$
  • $\begingroup$ The Dirac Delta is not a function and those symbols are not integrals. And inasmuch as the Dirac Delta is not a function, it has no meaning to write $\delta(x)=0$ for $x\ne $0$. $\endgroup$ – Mark Viola Mar 27 '18 at 17:11
  • 1
    $\begingroup$ See This Answer, This Answer, and This Answer, in which I provided some primers on the Dirac Delta. $\endgroup$ – Mark Viola Mar 27 '18 at 17:28
  • $\begingroup$ This is one of the two axioms of the Dirac delta function. The other is that $\delta(x) =0$ when $x \not = 0$. $\endgroup$ – Carl Mummert Mar 31 '18 at 19:31
2
$\begingroup$

Usually a way to get a feeling about this $\delta(x)$ "function" is to consider a sequence of functions $\delta_n(x)$ such that $$\lim_n\delta_n(x)=\delta(x)= \begin{cases} 0,&x\ne 0\\\\ \infty,&x=0 \end{cases} $$ For example a sequence of rectangle functions $\delta_n(x):=n$ for $x\in[-1/2n,1/2n]$ and $\delta_n(x)=0$ for $x\notin[-1/2n,1/2n]$ would work. The pointwise limit of this sequence is indeed $\lim_n\delta_n(x)=\delta(x)$. For each $n$ we have $$\int^{\infty}_{-\infty}\delta_n(x)\,dx=\int^{1/2n}_{-1/2n}n\,dx=1$$ So the integral makes sense and it is well defined for each element of the sequence $\{\delta_n(x)\}$. However if one is to interpret the "integral" $$\int^{\infty}_{-\infty}\delta(x)\,dx$$ in the usual sense of Lebesgue for instance this integral would normally be zero since $\delta(x)$ is zero except at a single point $x=0$. In the language of measure theory a single point turns out to be a set of measure zero (a single point has no length unlike an interval) thus the contribution to integral at this single point is zero. However the "integral" above is defined to actually represent the following $$\int^{\infty}_{-\infty}\delta(x)\,dx=\lim_n\int^{\infty}_{-\infty}\delta_n(x)\,dx=1$$ Therefore as mentioned in comments above this "integral" should not be interpreted as an integral in the usual sense. And of course such a "function" $\delta(x)$ is not a proper function.

$\endgroup$
13
$\begingroup$

Nobody can prove that from that sort of definition. The Wikipedia article that you mention states clearly that what you wrote “is merely a heuristic characterization. The Dirac delta is not a function in the traditional sense as no function defined on the real numbers has these properties” (emphasis added).

$\endgroup$
5
$\begingroup$

Let us look at the distribution sense that \begin{align*} \left<\widehat{\delta},\varphi\right>&=\left<\delta,\widehat{\varphi}\right>\\ &=\widehat{\varphi}(0)\\ &=\int_{{\bf{R}}^{n}}\varphi(x)dx\\ &=\left<1,\varphi\right>, \end{align*} so in some sense we have $\widehat{\delta}=1$, if we view that $\widehat{\delta}(0)$ as the integral of $\delta$, then $\displaystyle\int\delta(x)dx=1$.

Note that for a regular function $\varphi$, we have $\widehat{\varphi}(0)=\displaystyle\int_{{\bf{R}}^{n}}\varphi(x)e^{-2\pi i\cdot 0}dx=\displaystyle\int_{{\bf{R}}^{n}}\varphi(x)dx$.

$\endgroup$
  • 9
    $\begingroup$ Based on the OP and the lack of details in this answer, I am deeply skeptical about its usefulness. For someone who does not know already about distributions (and someone who does would not have asked this question in the first place), it is utterly unintelligible. $\endgroup$ – Clement C. Mar 27 '18 at 17:11
  • $\begingroup$ Yes, I agree... Sorry about that. $\endgroup$ – user284331 Mar 27 '18 at 17:12
  • $\begingroup$ No need to apologize! But tayloring the answers to the OP is a good thing to keep in mind. $\endgroup$ – Clement C. Mar 27 '18 at 17:13
  • $\begingroup$ There are many heuristic definitions of the $\delta$-function as limits of proper functions (e.g., normalized Gaussian with vanishing variance), where the integral can be computed. $\endgroup$ – David G. Stork Mar 27 '18 at 17:13
  • 1
    $\begingroup$ I think this sort of question is probably searched a lot on the internet, so keeping this around will almost certainly be useful for someone, which is why I am upvoting. $\endgroup$ – Theo Diamantakis Mar 27 '18 at 17:37
3
$\begingroup$

You cannot prove this, for the reason that $\delta$ so defined is not a well-defined object. For instance, it is not a well-defined function, hence cannot be integrated as a function.

The part $\int\limits_{-\infty}^{\infty}\delta(x)dx=1$ is usually taken to be part of the (non-rigorous) definition of $\delta$. Using this, and assuming that some usual rules about integral calculus still hold, one can "prove" that $\int\limits_{-\infty}^{\infty}f(x)\delta(x)dx = f(0)$.

The proper definition of $\delta$ requires some more machinery. Essentially, $\delta$ is said to be the functional on some function space, that acts as $\delta: f \mapsto f(0)$. That is, this functional maps a function $f$ to its value at zero $f(0)$. Such objects are also called distributions or generalized functions, and you can read more about them here.

$\endgroup$
  • 3
    $\begingroup$ Actually in measure theory we talking about functions $f:X\to[0,\infty]$ all the time; they have well-defined integrals (if they're measurable, of course). So the bad bad bad definition he gives does define a well-defined function. But it doesn't define the Dirac delta - if $\delta$ is as he defines it then $\int\delta=0$. $\endgroup$ – David C. Ullrich Mar 27 '18 at 17:09
2
$\begingroup$

Recall that we can change the value of a function at one point and the integral won't change. Thus, using your definition of the Dirac Delta, we have that $\int_{\mathbb{R}} \delta(t) dt = \int_{\mathbb{R}} 0 dt = 0$.

So where is the issue? It lies in that the Dirac Delta is not actually a function; instead, the Dirac Delta is a distribution. In some sense, distributions generalize properties of functions. Note though that we can view the Dirac Delta as a limit of a sequence of functions defined so as to shrink towards $0$ everywhere but in a small enough neighborhood of the origin, in which a large spike occurs.

If you're unfamiliar, imagine the graph of a normal distribution centered at the origin with very low variance.

$\endgroup$
2
$\begingroup$

If we view the $\delta$ as the limit of something called "nascent" delta functions (which I think can be understood just fine in this context without much advanced math).

For any (positive for convenience) $\eta$ with $\int_\mathbb{R}\eta=1$ on $\mathbb{R}$, we define $$ \eta_\epsilon=\frac{1}{\epsilon}\eta\left(\frac{x}{\epsilon}\right) $$ which still integrates to $1$ (by u substitution).

Then, $$ \lim_{\epsilon\to 0}\int_\mathbb{R}\eta_\epsilon(x-y)f(y)\mathrm dy=f(x) $$ i.e. these nascent deltas behave like the dirac delta when $\epsilon>0$ is small; they forget parts of $f$ far away from $x$, and remember things better and better as you get closer to $x$.

Examples of the above:

The heat kernel/Gaussian, $$ \eta_\epsilon(x)=\frac{1}{\sqrt{2\pi \epsilon}}\exp\left(\frac{-x^2}{2\epsilon}\right) $$ or a very nice example where we use a sequence of integers to approximate the delta at $0$, $$ \eta_n(x)=\begin{cases}\frac{n}{2}\cos(nx)& |x|<\frac{\pi}{2n}\\ 0&\text{otherwise} \end{cases} $$ where we have $$ \lim_{n\to \infty}\frac{n}{2}\int_{-\frac{\pi}{2n}}^{\frac{\pi}{2n}}\eta_n(y)f(y)\mathrm dy=f(0) $$ for $f$ nice (it's easy to see for smooth $f$ using IBP, still not so bad to prove for $f$ continuous). I like this one, since you can really see the approximation happening explicitly.

Anyway, to your question, with this definition becomes not so bad, since taking $f\equiv 1$, we have $$ \int_{\mathbb{R}}\delta(x)\mathrm dx=\lim_{\epsilon\to 0}\int_\mathbb{R}\eta_{\epsilon}(x-y)f(y)\mathrm dy= \lim_{\epsilon\to 0}\int_\mathbb{R}\eta_{\epsilon}(x-y)\mathrm dy=1 $$ since the integral of each $\eta_\epsilon=1$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.