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

Question

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!

Answer 1

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).

Answer 2

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 $\widehat{\varphi}(0)=\displaystyle\int_{{\bf{R}}^{n}}\varphi(x)e^{-2\pi i\cdot 0}dx=\displaystyle\int_{{\bf{R}}^{n}}\varphi(x)dx$.

Answer 3

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.

Answer 4

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 $\eta_\epsilon=1$.

Answer 5

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.

Answer 6

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.

Answer 7

Actually, one can prove this result in a rigorous way, and the answer is due to the fact that one can identify measures with distribution of order one. One can define the Dirac delta as a distribution of order one defined for any $\varphi\in C^\infty_c = \mathcal{D}$ by $$ \langle\delta_0,\varphi\rangle_{\mathcal{D}',\mathcal{D}} = \varphi(0) $$ but also as a measure, defined for any set $A⊂\mathbb{R}^d$ by $$ \tilde{\delta}_0(A) = 1 \text{ if } 0∈ A, \\ \tilde{\delta}_0(A) = 0 \text{ if } 0\notin A. $$ (where I use the notation $\tilde{\delta}_0$ to indicate that for now I did not prove how can one identify them). Then, integration with respect to a measure is well defined. Moreover, one can write for any measure $μ$ and any $\varphi ∈C^0_0$ $$ \langle\mu, \varphi\rangle_{\mathcal{M},C^0_0} =\int_{\mathbb{R}^d} \varphi(x)\,\mu(\mathrm{d}x), $$ defining a linear functional $\mu\in \mathcal{M} = (C^0_0)'$, the set of bounded radon measures. This is the content of Riesz representation theorem (this is similar to the fact that $L^1$ functions can be seen as distributions), and one can then easily identify $\mu$ with a distribution by restricting the set of test function to $C^\infty_c$ instead of $C^0_0$. Therefore, $\tilde{\delta}_0$ defines a distribution defined by $$ \langle\tilde{\delta}_0, \varphi\rangle_{\mathcal{D}',\mathcal{D}} = \int_{\mathbb{R}^d} \varphi(x) \,\tilde{\delta}_0(\mathrm{d}x). $$

• To see that $\tilde{\delta}_0 = \delta_0$, with the above identification, remark that by taking a sequence of nice functions $\varphi_n$ converging to $\mathbb{1}_A$ for some set $A$, one gets $$ \langle\delta_0,\varphi_n\rangle_{\mathcal{D}',\mathcal{D}} = \varphi_n(0) \underset{n\to\infty}{\to} \tilde{\delta}_0(A). $$ Therefore, it is correct to write $\int_{\mathbb{R}^d} \,\delta_0(\mathrm{d}x)$

• Now we can prove the result, just by writing that $$ \int_{\mathbb{R}^d} \,\delta_0(\mathrm{d}x) = \delta_0(\mathbb{R}^d) = 1 $$