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1) Can we solve if we have two delta functions with $a=b$? Can we solve this? If not why?

$$\int\delta (x−a)\delta(a−x)dx$$

2) Can we solve this

$$\int f(x)\delta (x)dx$$

...where $f(x)=1/x$ i.e. $f(x)$ is infinite at $x=0$ and also delta function is infinite at $x=0$. We can have many such similar $f(x)$ which may face same difficulty.

user224528
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    Wrong title and lead—there's no equation to solve here, it's about integrals to calculate... – CiaPan Mar 18 '15 at 12:08

3 Answers3

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The problem you are facing with this lies in the fact that the Dirac-delta distribution is not actually a function, it's a type of measure. So you can't treat it formally in the same sense you would treat a function $f:\mathbb{R} \to \mathbb{R}$. When you integrate a function with respect to normal integration

$$ \int f(x) dx $$

you are implicitly integrating the function $f$ with respect to a type of measure called Lebesgue measure, and Lebesgue measure is for the most part consistent with Riemann integration (they are exactly the same in the case of continuous functions). When you write $\int f(x) \delta(x-a) dx$ it looks like you're doing Riemann integration, but you are in fact not. Instead you are integrating the function $f$ with respect to a different measure.

Another way of thinking about measure, is to see measure as assigning "mass" or importance to certain parts of the real line. In Lebesgue measure, all points are treated equally and mass is assigned to intervals and corresponds to length as we normally would interpret it. For the Dirac measure given by $\delta(x-a)$, one point is considered to contain all of the mass, namely the point $a$.

So when we think of integrating with respect to Dirac measure, it doesn't really make sense to treat $\delta(x-a)$ as a function. Therefore writing $\int \delta(x-a) \delta(a-x)dx$ doesn't really make sense. Another important point is that given a measure defined on a set, we can describe functions as being measurable with respect to that certain measure. It is only when a function is measurable with respect to a certain measure that we are then allowed to integrate it with respect to that measure. It so turns out that $f(x) = \frac{1}{x}$ is not Dirac-measurable, therefore we can't compute the integral $\int f(x) \delta(x)$.

Mnifldz
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  • This means ∫f(x)δ(x)dx is not equal to f(0) if f(x)=1/x or of similar nature. Thx for the reply.. – user224528 Mar 18 '15 at 08:06
  • @user224528 Yes that is correct. In order to integrate a function $f$ with respect to the $\delta$-measure, it needs to first be shown that $f$ is $\delta$-measurable. If you're interested in learning more about this area of math, I would search for books on articles on "measure theory" (surprise). – Mnifldz Mar 18 '15 at 08:09
  • We study f(x)=1/x. It is (+Infinity) around x=0 if x is very small but positive and It is (- Infinity) around x=0 if x is very small but negative. That is, along +X axis f(x) is positive and along -X axis f(x) is negative. Delta function peaks at x=0 but always positive all along X axis. Then summation of f(x)δ(x) for all X is zero. Leading to ∫f(x)δ(x)=0. What is the problem in this logic? – user224528 Mar 18 '15 at 09:24
  • Leading to ∫f(x)δ(x−a)dx=0. Small correction as I had forgotten to mention dx in integration. – user224528 Mar 18 '15 at 09:25
  • Leading to ∫f(x)δ(x)dx=0 for f(x)=1/x. Sorry for earlier statement. – user224528 Mar 18 '15 at 09:57
  • @user224528 I'm guessing that your physics or calculus professor told you to think of $\delta(x)$ as some infinitely high and narrow spike at $x=0$, but this is actually not a proper formal way to think of it. Your argument may be appropriate for the sake of integrating $f$ times some increasingly narrow function at $x=0$, but to do it properly we would have to take limits of integral of some functions that do this (maybe Gaussians). If we take this approach we may run into problems of convergence however. Is there a specific application of this integral you're using? Anything published? – Mnifldz Mar 18 '15 at 14:40
  • I'd say the most formally appropriate term is "generalized function" – Vlad Apr 29 '15 at 03:24
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Not exactly but the mathematical difficulty faced in the above integration (with a delta function) has an impact on the traditional interpretation of the concept of radiation by a point charge (Larmor formula) in Physics.

user473139
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The papers in Canadian Journal of Physics: A relook at radiation by a point charge- I and II are based on above discussion. In conventional Poynting’s theorem analysis, the rate of change of work done on a system of charges is written as addition of two terms, rate of change of stored energy and surface integral of Poynting vector. For a delta function source, the first two terms of this equation are either non-integrable or difficult to evaluate.

user473139
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