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First and foremost, apologies in advance for using an abuse of notation by placing the Dirac measure inside an integral for which I was told that this should not be done from a previous question asked by me. But given the circumstances, I have no choice.

This is essentially a word by word copy of an interpretation given on page 1 of these Berkeley notes:

The important property of the delta function is the following relation $$\displaystyle\int f(t) \delta(t) \, \mathrm{d}t = f(0)$$ for any function $f(t)$. This is easy to see. First of all, $\delta(t)$ vanishes everywhere except when $t = 0$. Therefore, it does not matter what values the function $f(t)$ takes except at $t = 0$. You can then say $f(t)\delta(t) = f(0)\delta(t)$. Then $f(0)$ can be pulled outside the integral because it does not depend on $t$, and you obtain the r.h.s.

Here's the problem, it was my understanding that $$\delta(t) = \begin{cases} 0 & \space \mathrm{for} \space t \ne 0 \\\infty&\ \mathrm{for} \space t = 0 \end{cases} $$ So by my logic this means that $\delta(0)=\infty$ and therefore undefined; which implies that when $t=0$ $$\displaystyle\int f(0) \delta(0) \, \mathrm{d}t = \displaystyle\int f(0) (\infty) \, \mathrm{d}t$$ which is manifestly not true and certainly not equal to $f(0)$.

Clearly I am missing the point of this argument, so if someone would be kind enough to explain it to me I would be most grateful.

Thank you in advance.

Regards,

Blaze

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BLAZE
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    Your logic is correct : the definition you use is flawed from a mathematical standpoint, so you get an absurdity. This definition can be usefull in physic, but I advise to use instead a correct mathematical definition. – Tryss Oct 05 '15 at 11:54
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    just use the linear functional interpretation and you will never have such trouble $( \delta(x),f(x) )=f(0)$ by definition – tired Oct 05 '15 at 11:57
  • @Tryss Would you care to tell me this "correct mathematical definition"? – BLAZE Oct 05 '15 at 12:04
  • @tired Please tell me this "linear functional interpretation" – BLAZE Oct 05 '15 at 12:09
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    @BLAZE : the general idea is to define $\delta$ by $\delta(f) = f(0)$. The domain of $\delta$ is then a space of functions instead of $\Bbb R$. – Tryss Oct 05 '15 at 12:15
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    the wikipedia article about "distributions" will tell you a lot... – tired Oct 05 '15 at 12:17
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    as alternative, look up a concrete representation of the delta function $\delta(x)=\lim_{\epsilon->0}\psi(x,\epsilon)$ and consider their action on an arbritary (Schwartz)- function$\lim_{\epsilon->0}\int_{\mathbb{R}} \psi(x,\epsilon)f(x)$ – tired Oct 05 '15 at 12:25

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The first time I encountered the Dirac delta was when I did my first (and only) course in electrostatics. It was introduced in the following way

$$ \delta(x) = \begin{cases} 0 & \space \mathrm{for} \space x \ne 0 \\\infty&\ \mathrm{for} \space x = 0 \end{cases} $$

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

We were told that we should think of this as an enormous high spike at $x=0$ which one might construct by taking the limit of rectangles with height $n$ and width $1/n$. The interesting property of the Dirac delta is the following

$$ f(x) \delta(x) = f(0)\delta(x) $$

for all "ordinary functions" i.e. not another delta function (sometimes limited to continuous functions). This should be clear since $f(x)\delta(x)$ has to be $0$ when $x \neq 0$. Now we can use this fact on the integral

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

and since $f(0)$ is a constant

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

I hope this makes some intuitive sense to you.

Now I don't like this explanation since I've been given a more rigourous one in a recent course on measure theory and the Lebesgue integral. Here the Dirac delta can be defined as a measure on a set in a much nicer way that maintains this property when integrating.

Lundborg
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  • Thanks for your answer; you have substituted $x=0$ for $f(x)$ but not for $\delta(x)$ giving $\displaystyle\int_{-\infty}^{\infty} f(0) \delta(x) dx = f(0) \int_{-\infty}^{\infty} \delta(x) dx = f(0)$ how can you justify this? – BLAZE Oct 05 '15 at 12:15
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    i think $"f(x)\delta(x)=f(0)\delta(x)"$ is not really correct in a formal way...it's useful for calculations if you know how to handle it, but nothing more... – tired Oct 05 '15 at 12:21
  • @tired Yes but you can't substitute $x=0$ for $f(x)$ and then not for $\delta(x)$ it makes no sense whatsoever. – BLAZE Oct 05 '15 at 12:33
  • @Neutronic "Here the Dirac delta can be defined as a measure on a set in a much nicer way that maintains this property when integrating." What property are you referring to? – BLAZE Oct 05 '15 at 12:36
  • @BLAZE if you keep in mind that you formally have to apply this expression on a test function it makes kind of sense. but it is not rigorous in any way – tired Oct 05 '15 at 12:58
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    The property im referring to is the one mentioned at the end, the fact that $$\int_{-\infty}^{\infty} f(0) \delta(x) dx = f(0) \int_{-\infty}^{\infty} \delta(x) dx = f(0)$$ BLAZE, can you not see that $f(x)\delta(x) = f(0)\delta(x)$ ? – Lundborg Oct 05 '15 at 13:04
  • @Neutronic No sorry I can't see why $f(x)\delta(x)=f(0)\delta(x)$, that is the root of my misunderstanding I think, so please elaborate, thanks – BLAZE Oct 05 '15 at 13:21
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    What you need to think of is that $f(x)\delta(x)$ is a pointwise product of two functions. For each $x$ we take the product of $f(x)$ and times it by $\delta(x)$. Now, since we know that for all $x \neq 0$ we have that $\delta(x) = 0$, the values of $f(x)$ are irrelevant, $f(x)$ can take any (reasonable) value but the product $f(x)\delta(x)$ will still be $0$. So we might aswell just write $f(0)\delta(x)$. This is still a function of $x$ that works for the same $x$ as the product function. – Lundborg Oct 05 '15 at 13:35
  • @Neutronic What is "pointwise product of two functions" and why is it that the values of $f(x)$ are irrelevant? – BLAZE Oct 05 '15 at 13:56
  • Pointwise means that you start by picking an $x$ and then do your product with that $x$ inserted into each function. For instance with the functions $g(x)=2x$ and $h(x)=x^2$ we would calculate the pointwise product for $x=2$ by saying $g(2)=4$ and $h(2)=4$ so $g(2)h(2)=4*4=16$. A uniform (the opposite of pointwise) way of doing this would be by multiplying the functions first i.e. $g(x)h(x) = 2x(x^2)=2x^3$ and then inserting $x$ which in this case leads to the same result.

    The values of $f$ for $x \neq 0$ are irrelevant in the product $f(x)\delta(x)$ since $\delta(x)=0$ for the same $x$.

     – Lundborg Oct 05 '15 at 14:04
  • @Neutronic Okay so one more question, you've been a great help so far but $\delta(0)=\infty$ however you 'dress it up' and you just said yourself that $g(2)h(2) = 4 \times 4=16$ yet you claim that $f(0)\delta(0) \ne f(0)\infty$? If this is not fully justified I will have to start another post on it I'm afraid, sorry. – BLAZE Oct 05 '15 at 14:25
  • At no point have we attempted to evaluate $f(x)\delta(x)$ in $x=0$. You will never see a $\delta(x)$ outside of an integral unless you made a mistake. If we were to do so then yes we would get $\infty$. – Lundborg Oct 05 '15 at 14:29
  • @Neutronic okay, now we are getting somewhere, you say "At no point have we attempted to evaluate $f(x)\delta(x)$ in $x=0$" but I thought that was exactly what we are doing, can you give a reason why not? Also use @{name} at the start to make sure I see the message. – BLAZE Oct 05 '15 at 14:42
  • @BLAZE the reason is we don't want to deal with infinities. The point of the delta function is to "pick out" the value at zero under the integral. Again, as tired already pointed out, the delta function is not a very rigorous piece of math and its often only used by physicists. I would consider asking this question on their stack exchange, since they might be better at dealing with some of the conflicts that arise when using it. – Lundborg Oct 05 '15 at 14:57
  • @Neutronic "under the integral" means in the integrand right? – BLAZE Oct 05 '15 at 15:00
  • Yes it does. Language :) – Lundborg Oct 05 '15 at 15:02