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I have been searching for a derivation of the defining property for the Dirac-delta function: $\displaystyle\int_{x=-\infty}^{x=\infty} f(x) \delta(x) \, \mathrm{d}x = f(0)$ and found this derivation on the first 2 pages:

Defining the Heaviside (step) function $H(x)$ as

$$H(x) = \begin{cases} 0 & \text{for } x \lt 0 \\1& \text{for } x \gt 0 \end{cases} $$ 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

$$\delta(x) = \begin{cases} 0 & \text{for } x \ne 0 \\\infty& \text{for } x = 0 \end{cases} $$ 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 = \color{green}{\left.\vphantom{\frac 1 1} f(x)H(x) \right|_{x=-\infty}^{x=\infty}} - \int_{\color{red}{x=-\infty}}^{x=\infty} f^\prime(x)H(x) \, \mathrm{d}x \\[10pt] = {} &0-\int_{\color{red}{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}

Could someone please explain how the limits marked $\color{red}{\mathrm{red}}$ were changed?

Thank you.

(and yes I know from last time, it's an abuse of notation placing the Dirac measure/distribution inside an integral)

Michael Hardy
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BLAZE
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2 Answers2

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The way the limits in red were changed is simply that the piecewise definition of $H$, stated earlier in the question, says that $H(x)=0$ when $x<0$. Thus $$ \int_{-\infty}^0 (\text{anything}\times H(x))\, dx = 0. $$

The idea that $\delta(0)=\infty$ should not be taken too literally. Notice that $$ \int_{-\infty}^\infty 3.4\delta(x) f(x)\,dx = 3.4f(0), $$ so one would then say that this "infinity" is $3.4$ times as big as the earlier "infinity". But no attempt is made to give such concepts any precise definition. If one wants to show that $H'(x)=\delta(x)$, one can observe that $H$ has a vertical jump at $0$ and therefore say there is an infinite slope, but again no attempt is made to make that precise. Rather, one precisely defines $H'$ by integrating by parts, as you did.

Michael Hardy
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    You beat me to it! +1 – Mark Viola Oct 04 '15 at 02:36
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    +1 For the comments - that is not really a "proof", rather a motivation. Besides the property $\int_{x=-\infty}^{x=\infty} f(x) \delta(x) , \mathrm{d}x = f(0)$ does not require that the function vanishes at $\infty$ – leonbloy Oct 04 '15 at 02:44
  • @leonbloy Do you know of such a proof/derivation, there must be one otherwise it would not exist? – BLAZE Oct 04 '15 at 02:47
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    The proofs were first given by the French mathematician Laurent Schartz in the 1940s. There are a number of books on the subject, including Theory of Distributions by Richards and Youn. ${}\qquad{}$ – Michael Hardy Oct 04 '15 at 02:50
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    @BLAZE It's not clear for me what you want to prove. If you want to formally prove the propery of the title, then you first need a formal definition of the Dirac "function" (which actually is not a function). To define it as the derivative of $H(x)$ does not fit the bill, as $H(x)$ is not (really) derivable at $x=0$. For informal uses, the property of the title is not taken as a property to prove, but rather as part of the Dirac delta definition. – leonbloy Oct 04 '15 at 02:54
  • @Michael Thank you for your very swift answer. But does this mean that we cannot write ${\left.\vphantom{\frac 1 1} f(x)H(x) \right|_{x=-\infty}^{x=\infty}}$ marked $\color{green}{\mathrm{green}}$ in my question? – BLAZE Oct 04 '15 at 02:57
  • The thing in green is fine. It is equal to $0$ because $f(x)\to0$ as $x\to\pm\infty$. ${}\qquad{}$ – Michael Hardy Oct 04 '15 at 03:01
  • @leonbloy so is this wrong? If I changed the Title and asked you to formally derive/prove (as Dirac did) the property in the title, could you do it? – BLAZE Oct 04 '15 at 03:02
  • @Michael Sorry, what I was getting at is, should $\color{green}{\left.\vphantom{\frac 1 1} f(x)H(x) \right|{x=-\infty}^{x=\infty}}$ really be written as $\color{green}{\left.\vphantom{\frac 1 1} f(x)H(x) \right|{x=0}^{x=\infty}}$? – BLAZE Oct 04 '15 at 03:06
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    @BLAZE Yes, the treatise of the embedded link contains non-sense. – Mark Viola Oct 04 '15 at 03:08
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    @BLAZE : No. There would be issues about $f(x)H(x)$ approaching $0$ from the left or the right, and functions other than well-behaved test functions or Schwartz functions like $f(x)$ aren't really taken to be defined at individual points in this subject. ${}\qquad{}$ – Michael Hardy Oct 04 '15 at 03:09
  • @Dr.MV Just to confirm. Is this really non-sense? – BLAZE Oct 04 '15 at 03:11
  • @BLAZE Yes, it is. The Dirac Delta in not a function. Therefore, writing $\delta (x)=0$ for $x\ne 0$ and $\delta (x)=\infty$ at $x=0$ is absolutely silly. However, the Dirac Delta is explained rigorously as a Generalized Function, also known as Distribution Theory. – Mark Viola Oct 04 '15 at 03:15
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    @BLAZE : I wouldn't use the word "nonsense" but I would say the definition of the $\delta$ function stated on the first page should not be taken literally. Perhaps one should say that to take it literally would be nonsense, but if you know it's not meant literally, then it might not be nonsense. ${}\qquad{}$ – Michael Hardy Oct 04 '15 at 03:15
  • @MichaelHardy When you said "There would be issues about $f(x)H(x)$" were you referring to $f(x)$ in particular? – BLAZE Oct 04 '15 at 03:19
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    @BLAZE : No. $f(x)$ would be a very well behaved function, but $H(x)$ behaves badly at $0$. ${}\qquad{}$ – Michael Hardy Oct 04 '15 at 03:39
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    @everybody: This is the bill we have to pay, abusing the notation using the integral symbol. Indeed, it is handy if you know how things work, but before you do, it is just a mess. Btw, I wonder who the cited Sir Issac Newton was... – mickep Oct 04 '15 at 04:02
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    @mickep when i see all this questions, i am really happy that at some point i decided to learn distribution theory properly (at least at a level sufficent for a physicist) and i would be really a big help if this topic would be part of the standard curriculum ... – tired Oct 04 '15 at 12:06
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Probably not as much rigorous: $$\int_{-\infty}^{+\infty}f(x)\delta(x)\,dx=\int_{-\infty}^{+\infty}(f(x)-f(0))\delta(x)\,dx+\int_{-\infty}^{+\infty}f(0)\delta(x)\,dx\\=\int_{-\infty}^{+\infty}(f(x)-f(0))\delta(x)\,dx+f(0)\int_{-\infty}^{+\infty}\delta(x)\,dx\\=\int_{-\infty}^{+\infty}(f(x)-f(0))\delta(x)\,dx+f(0)$$ by definition of $\delta(x)$ it follows that $(f(x)-f(0))\delta(x)=0$ for all $x\in(-\infty,+\infty)$. So the last integral vanishes and we are left with $$\int_{-\infty}^{+\infty}f(x)\delta(x)\,dx=f(0)$$

Arian
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