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I understand that $\delta(x)=0$ whenever $x \ne 0$ and that $\displaystyle\int_{x=-a}^{x=b} \delta(x) \, \mathrm{d}x = 1 \space$ $\forall\, a,b \gt 0$ and also that $\displaystyle\int_{x=-\infty}^{x=\infty} \delta(x-a) \, \mathrm{d}x = 1$.

But I see no justification that $\color{blue}{\displaystyle\int_{x=-\infty}^{x=\infty} f(x) \delta(x) \, \mathrm{d}x = f(0)}$ for any arbitrary function $f(x)$.

I'm asking this question because the formula marked blue was given to me in response to this previous related question asked by me. But every-time I search the internet for an explanation of its derivation all I get is the same formula stated without proof. Hence, could someone please prove and/or explain the origin of the formula marked blue?

Thank you.

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    $\begingroup$ Your formula in blue is the usual definition of what $\delta$ means. It implies your earlier relation (take $f$ as the constant function, $1$). $\endgroup$ – Simon S Oct 1 '15 at 19:19
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    $\begingroup$ The Dirac measure (Dirac distribution) is defined as $\delta[f] = f(0)$ (for suitably regular $f$). Writing it in an integral is just an abuse of notation. $\endgroup$ – Daniel Fischer Oct 1 '15 at 19:20
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    $\begingroup$ if $\delta(x)=0$ whenever $x \ne 0$ then $\int_{x=-a}^{x=b} \delta(x) \, \mathrm{d}x = 0$, right? Not 1. $\endgroup$ – Guillermo Mosse Oct 1 '15 at 19:21
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    $\begingroup$ @SeñorBilly Whatever else $\delta$ is, it is not a Riemann or Lebesgue integrable function. More properly, Daniel's notation suggests what $\delta$ is: a function from some appropriately defined set of functions ($\mathbb R \to \mathbb R$) to the set of the real numbers. That said, the integral notation is a useful abuse of notation. $\endgroup$ – Simon S Oct 1 '15 at 19:23
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    $\begingroup$ I know that you didn't come up with it yourself. And although I strongly dislike this particular abuse of notation, since it creates more confusion than it helps, there is nothing wrong eo ipso with abuse of notation. Some abuses of notation help a lot. $\endgroup$ – Daniel Fischer Oct 1 '15 at 20:21
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In most common definitions of the Dirac delta (generalized) function, the formula in your post is taken as a definition. But if, for example, you accept that $\delta(x)$ is the Fourier transform of $1$, then it can be proven as follows:

$$\int_{-\infty}^{\infty} f(x) \delta(x) dx = \int_{-\infty}^{\infty} f(x) \left( \int_{-\infty}^{\infty} e^{-2\pi i kx} dk \right) dx = $$

$$= \int_{-\infty}^{\infty} \left( \int_{-\infty}^{\infty} f(x) e^{-2\pi i kx} dx \right) dk = \int_{-\infty}^{\infty} \tilde{f}(k) dk = \int_{-\infty}^{\infty} \tilde{f}(k) e^{2\pi i k\cdot 0} dk = f(0)$$

where $\tilde{f}(k)$ is the Fourier transform of $f(x)$, and we assume the function is nice enough that we can change the order of integration.

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You can define the delta function, as folows: Define $$ \delta_a(x) = \begin{cases} 1/a & |x|\leq a/2\\ 0 & \text{otherwise.} \end{cases} $$ So we see that this function defines a box of width $a$ and height $1/a$, so that $$ \int_{\mathbb{R}} \delta_a(x)\;dx=1,\quad\forall a\in\mathbb{R}_{>0}. $$ We can now think of taking the limit as $a\to0⁺$, whereas our box tends to something infinitely thin and infinitely tall but still with total area $1$. In that case we also have $$ \lim_{a\to\,0⁺}\delta_a(x) = 0, \quad\forall x\neq0. $$ Let's take the condition $$ \int_{\mathbb{R}} \delta(x)f(x)\;dx = f(0) $$ as the definition of the delta function, and let's see whether $\lim_{a\to0⁺}\delta_a(x)=\delta(x)$. For any (integrable) $f$ we have $$ \int_{-\infty}^\infty \delta_a(x)f(x)\;dx=\int_{-a/2}^{a/2} \delta_a(x)f(x)\;dx=\frac{1}{a}\int_{-a/2}^{a/2} f(x)\;dx, $$ Where we have used that $\delta_a(x)=1/a$ on $[-a/2,a/2]$ and $0$ elsewhere. Now by the mean-value theorem we have that there exists a number $\mu(a)\in[-a/2,a/2]$, such that $f(\mu(a))$ is the mean-value of $f$ on $[-a/2,a/2]$. The result is that we can write $$ \int_{-a/2}^{a/2} f(x)\;dx=f\big(\mu(a)\big)\int_{-a/2}^{a/2}dx= af\big(\mu(a)\big), $$ which in turn gives us $$ \int_{-\infty}^\infty \delta_a(x)f(x)\;dx=f\big(\mu(a)\big). $$ Taking the limit as $a\to0⁺$, we see that $\mu(a)$, being confined to the interval $[-a/2,a/2]$, which tends to the singleton $\{0\}$ as $a$ tends to $0⁺$, must also go to $0$, and hence, assuming continuity of $f$, we have $$ \int_{-\infty}^\infty \lim_{a\to\,0⁺}\,\delta_a(x)f(x)\;dx=\lim_{a\to\,0⁺}\int_{-\infty}^\infty \delta_a(x)f(x)\;dx=\lim_{a\to0⁺}f\big(\mu(a)\big)=f(0). $$ As desired.

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  • $\begingroup$ It's usually preferred to take a smooth approximating sequence, so that you can do things like take derivatives by passing to the approximating sequence. $\endgroup$ – Ian Oct 10 '15 at 12:49
  • $\begingroup$ Well you're absolutely right. I just think that this "box" function approach is very intuitive, but there's a lot of different ways of defining the delta as the limit of a sequence of functions, for instance by gaussian distributions. $\endgroup$ – Erik Olesen Oct 10 '15 at 13:11
  • $\begingroup$ The point I was trying to make is just that, the dirac delta is accentually not a function, but rather an integral operator with the property, as mentioned above, that $\delta[f]=f(0)$, but if you for intuition want to think about it geometrically, you should consider an infinitely tall "spike" with total area 1, such that integrating $\delta(x)f(x)$ picks out exactly $f(0)$. $\endgroup$ – Erik Olesen Oct 10 '15 at 13:11
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Defining the Heaviside (step) function $H(x)$ as

$$H(x) = \begin{cases} 0 & \space \mathrm{for} \space x \lt 0 \\1&\ \mathrm{for} \space x \gt 0 \end{cases} $$ The derivative of the Heaviside function is zero for $x \ne 0$ and undefined for $x=0$ so the $\delta$ function can represent the derivative of the Heaviside function

$$\delta(x) = \begin{cases} 0 & \space \mathrm{for} \space x \ne 0 \\\infty&\ \mathrm{for} \space x = 0 \end{cases} $$ and $$\int_{x=-\infty}^{x=\infty}\delta(x)\mathrm{d}x=1$$

Let $f(x)$ be any continuous function that vanishes at $x=\pm\infty$ and integrating by parts

\begin{align} & \int_{x=-\infty}^{x=\infty} f(x)\delta(x) \, \mathrm{d}x = \left.\vphantom{\frac 1 1} f(x)H(x) \right|_{x=-\infty}^{x=\infty} - \int_{x=-\infty}^{x=\infty} f^\prime(x)H(x) \, \mathrm{d}x \\[10pt] = {} &0-\int_{x=0}^{x=\infty} f^\prime(x)H(x) \, \mathrm{d}x= \left.\vphantom{\frac 1 1}-f(x) \right|_{x=0}^{x=\infty}=f(0) \end{align}

QED

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    $\begingroup$ While this answer explains the formalism for computing with $\delta(x)$ as a distribution, your definition of $\delta(x)$ is anything but rigorous. $\endgroup$ – David Hill Oct 9 '15 at 17:32
  • $\begingroup$ @David Yes you're right, I acknowledge that, but this is as close as I can get it. Thanks for your input. $\endgroup$ – BLAZE Oct 9 '15 at 17:44
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    $\begingroup$ Just a note: The Heaviside [sic] function is named for Oliver Heaviside. :) $\endgroup$ – Andrew D. Hwang Oct 10 '15 at 12:57
  • $\begingroup$ @Andrew Thanks, changed it now :) $\endgroup$ – BLAZE Oct 10 '15 at 17:03

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