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Is the following equation mathematically rigorous? How can you tell?
## \int_{-\infty}^\infty \delta(x-a) \delta(x-b) dx=\delta(a-b)##
Thanks
## \int_{-\infty}^\infty \delta(x-a) \delta(x-b) dx=\delta(a-b)##
Thanks
What about the integral below?micromass said:It is not rigorous. You cannot multiply distributions like that. Things like ##\int_{-\infty}^{+\infty} \delta(x)^2 dx## have no meaning in the classical sense of distributions. While it is possible to extend distribution theory to include this case, this is by no means trivial (and these extensions are typically not very well-behaved).
Furthermore, something like ##\delta(a-b)## makes no sense outside an integral. And ##\delta(0) = +\infty## is most certainly incorrect.
Svein said:I have thought about my last step, and I have come up with this:
Since [itex]\int_{a-\epsilon}^{a+\epsilon}\delta(x-a) \delta(x-b)dx=0 [/itex] for |a-b|>ε, ε>0, we certainly have [itex]\int_{a-\frac{1}{n}}^{a+\frac{1}{n}}\delta(x-a) \delta(x-a-\frac{1}{n})dx=0 [/itex] for all n (just choose ε<1/2n). Thus [itex]\lim_{n\rightarrow\infty} \int_{a-\frac{1}{n}}^{a+\frac{1}{n}}\delta(x-a) \delta(x-a-\frac{1}{n})dx=0 [/itex] (just a tiny nagging doubt: Riemann integrals and limits are not necessary compatible).
But - I know that it is possible to construct a function ƒ∈C^{∞}, ƒ(x)∈[0, 1] with the following specification: Given real numbers a and b with a<b, ƒ(x)=0 for x≤a and ƒ(x)=1 for x≥b. And, of course, ƒ'(x)=0 for x<a and for x>b. Now, obviously ƒ'(x) will approach δ(x-a) as b→a.micromass said:The more nagging doubt should be that δ(x−a)δ(x−b)\delta(x-a)\delta(x-b) does not exist in the usual theory of distributions.
nuclearhead said:δ(x) = lim_{ε→0} 1/√ε exp(-x^{2}/ε)
pwsnafu said:Dirac is an equivalence class of said functions (specifically delta sequences), so you need to demonstrate your result is independent of the choice of the function. So no that's not rigorous either.
nuclearhead said:Why does that matter? You might say sin(x) is the equivalent class of all series definitions of sin(x). As long as one series converges you don't need to use the rest of them.
Shyan said:What about the integral below?
## \int_{-\infty}^\infty f(a) \left[ \int_{-\infty}^\infty \ \left( \int_{-\infty}^\infty e^{ix(k-a)} \ dx \right) \left( \int_{-\infty}^\infty e^{iy(k-b)} \ dy \right) \ dk \right] \ da ##
A rearrangement of the inner integrals gives ##\int_{-\infty}^\infty e^{ik(b-a)} \ dk##. I'm sure you're right about the product of distributions so it seems to me that either such a rearrangement or the equation ## \delta(x-a)=\int_{-\infty}^\infty e^{ik(x-a)} \ dk ## shouldn't be applicable here for some reason.
Lebesgue said:Hello! I'm interested in learning something about distributions and how to give rigorous meaning to the Dirac function. Does anyone know about any good textbook?
The Dirac delta formula, also known as the Dirac delta function, is a mathematical concept used to represent an infinitely narrow and infinitely tall spike at a specific point on a graph. It is often used in physics and engineering to model point masses or point charges.
The Dirac delta formula is important because it allows us to describe and analyze systems that involve point-like objects or impulses. It also has many applications in signal processing and differential equations.
The Dirac delta formula is rigorously defined as a distribution, rather than a function. This means that it cannot be evaluated at a single point, but instead must be integrated over a range. Mathematically, it is defined as a limit of a sequence of functions that approach an infinitely narrow spike at a specific point.
The Dirac delta formula has several important properties, including being infinitely tall at the point of evaluation, having a total area of 1 when integrated over all real numbers, and being odd and symmetric about the origin. It also follows the sifting property, which states that when integrated with a test function, it returns the value of the function at the point of evaluation.
No, the Dirac delta formula has limitations and should only be used in certain contexts. It is not a true function and cannot be evaluated at a single point, so it should not be used in traditional algebraic expressions. It is also not defined for non-real numbers, so it cannot be used in complex analysis.