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What is $\int_0^1 \delta(0)dx$? Should not it be equal to the number of reals in that interval?

My attempt:

$$\delta(x)=\frac2\pi\int_0^\infty \cos(xt)dt$$

(http://functions.wolfram.com/GeneralizedFunctions/DiracDelta/07/01/01/0004/)

$$\int_0^1 \delta(0) dx=\lim_{u\to 0}\frac2\pi\int_0^1\int_0^\infty \cos(uxt)dt dx$$

swapping integrals (do not know if justified):

$$\int_0^1 \delta(0) dx=\lim_{u\to 0}\frac2\pi\int_0^\infty \int_0^1 \cos(uxt)dx dt$$

$$\int_0^1 \delta(0) dx=\lim_{u\to 0}\frac2\pi\int_0^\infty \frac{\sin ut}{ut} dt$$

$$\int_0^1 \delta(0) dx=\lim_{u\to 0}\frac{ \text{sgn}(u)}{ u}$$

???

Another method gives

$$\left(\frac{\pi\delta(0)}2+\frac12\right)^{\frac{\pi\delta(0)}2-\frac12}$$

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Seen as a "function" $\delta(0)$ is not defined, so your integral is also not defined. There is no real number you can consistently assign to $\delta(0)$.

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  • $\begingroup$ What if seen as a distribution, not function? Dirac Delta is not a function, but distribution. $\endgroup$ – Anixx Jan 19 '17 at 0:30
  • $\begingroup$ @Anixx: it still doesn't have a value at zero. $\endgroup$ – Ross Millikan Jan 19 '17 at 0:33
  • $\begingroup$ I amended the question with my attempt at solving. $\endgroup$ – Anixx Jan 19 '17 at 0:37
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    $\begingroup$ @Anixx If seen as a distribution, you would need to apply the functional to the characteristic function of $[0,1]$ which isn't in the domain of any of the types of distribution that I'm familiar with. $\endgroup$ – Daniel Schepler Jan 19 '17 at 0:42
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You start with a representation of $\delta$, do some manipulations of formulas with complete disregard to their validity, and you end up with a limit that does not exist. In this sense you do actually get the correct result in the end: $\int_0^1 \delta(0){\rm d}x$ does not exist as a consistent mathematical expression, but there are many mistakes and misconceptions in this derivation so I would not have been surprised if you ended up with a finite result in the end. Let's try to go through it step by step.

The first thing that is sketchy here is that throughout this derivation you treat $\delta$ as it's a normal function. It's not. This is sometimes perfectly fine when doing calculations if you know what you are doing and the limitations of this. You overstepped these limitations here. Whenever you do this and end up with strange results then it's very likely that you have "broken the rules" at one or more steps.

The main mistake you have done is in taking $\delta(0) = \lim_{x\to 0} \delta(x)$. First of all $\delta$ is not continuous at $x=0$ which we would need for such an expression to be true and secondly there is no consistent value we can assign to $\delta(0)$ and have it satisfy the properties we want like $\int_{-\infty}^\infty \delta(x)f(x){\rm d}x = f(0)$ for "all" test functions $f(x)$.

You then integrate an expression that does not exist over $[0,1]$ and the result will be something that does not exist. However for arguments sake let's treat the right hand side of this as it was a legitimate expression.

Next you interchange the integration $\int_0^1{\rm d}x$ with the limit $\lim_{u\to 0}$ which cannot be justified in this case and next you interchange the two integrals $\int{\rm d}x\int {\rm d}t\to \int {\rm d}t\int{\rm d}x$ which you don't justify (see Fubini's theorem for conditions when this is allowed. At least you mention this which is something positive.

Then you correctly evaluate two integrals and finally you end up with a limit that does not exist. At least the "does not exist" is consistent thoughout the calculation.

The take away message here is that one should try to be very careful when doing calculations like this. The most important thing is to make sure the things you manipulate actually exists / are well defined in the first place and be careful to check that the manipulations you do are allowed (interchanging limits and integrals etc.).

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  • $\begingroup$ by the way, the both limits exist, $\lim_{x\to0}\delta(x)=0$ and $\lim_{x\to0} \frac{\operatorname{sgn} x} x=1$. Of course, my calculation is totally wrong, the limits are irrelevant here. $\endgroup$ – Anixx Jan 19 '17 at 1:50
  • $\begingroup$ I think the answer is either what is shown in the bottom of the question or simply $\delta(0)$. $\endgroup$ – Anixx Jan 19 '17 at 1:51
  • $\begingroup$ @Anixx No. $\frac{\text{sign}(x)}{x} = \frac{1}{|x|} \to \infty$ as $x\to 0$ does not exist. And my point for the other part was that $\lim_{x\to 0} \delta(x) \not = \delta(0)$ (this would be true if $\delta$ was a function and if it was continuous neither of which it is). $\endgroup$ – Winther Jan 19 '17 at 1:54
  • $\begingroup$ Oh, yes, the second limit does not exist (or rather, is infinite). The first one does. $\endgroup$ – Anixx Jan 19 '17 at 1:55
  • $\begingroup$ @Anixx If we treat $\delta(x)$ as a function satsfying $\delta(x) = 0$ for all $x\not = 0$ then we would have $\lim_{x\to 0}\delta(x) = 0$. However you cannot assign $\delta(0)$ to this value. You are assuming continuity by doing this. It's not continuous. $\endgroup$ – Winther Jan 19 '17 at 2:01
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Suppose we were to start over with setting up mathematics to do calculus, but now in a ultrafinitistic way; all calculations are done on finite lattices and the continuum limit is only taken at the end of the calculations. Then in an intermediary result before you take such a continuum limit, you can indeed give such an expression the interpretation of counting the number of points in the interval. But this then tends to infinity in the continuum limit.

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