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$$\int_{-\infty}^\infty f(x) \sqrt{\delta(x-a)} dx$$

How is this integral evaluated?

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- #1

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$$\int_{-\infty}^\infty f(x) \sqrt{\delta(x-a)} dx$$

How is this integral evaluated?

- #2

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- #3

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This is actually related to Bell's theorem. It is not the solution to the OP : searching A s.t. $$\int A (a,\lambda)A (\lambda,b)d\lambda=-cos (a-b) $$

If we make the assumption $$A (a,\lambda)=f (a-\lambda) $$ and let $$u=a-\lambda $$ we get a convolution :

$$\int f (u)f (a-b-u)du=-cos (a-b) $$

Using the Fourier convolution theorem :

$$2F^2 (\omega)=-\delta (\omega+1)-\delta (\omega-1)$$

Hence the searched function f is given by

$$f (x)=\frac {i}{\sqrt {2}}\int e^{ikx}\sqrt {\delta (k+1)+\delta (k-1)} $$

Bell's theorem shows that there is no such f.

If we make the assumption $$A (a,\lambda)=f (a-\lambda) $$ and let $$u=a-\lambda $$ we get a convolution :

$$\int f (u)f (a-b-u)du=-cos (a-b) $$

Using the Fourier convolution theorem :

$$2F^2 (\omega)=-\delta (\omega+1)-\delta (\omega-1)$$

Hence the searched function f is given by

$$f (x)=\frac {i}{\sqrt {2}}\int e^{ikx}\sqrt {\delta (k+1)+\delta (k-1)} $$

Bell's theorem shows that there is no such f.

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- #4

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Addendum : no such f in the range -1 to 1

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not every sequence of mathematical symbols makes sense∫∞−∞f(x)√δ(x−a)dx\int_{-\infty}^\infty f(x) \sqrt{\delta(x-a)} dx

How is this integral evaluated?

- #6

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You could aslo think it in the functional sense : ##\int\delta(x-y)f(y)dy=f(x)=:\Delta_x(f)##

Then the root of delta would be a semi evaluation of f , which you could compute by semi integrating, using Cauchy's formula for repeated integration extended to real values of fractional calculus :

##\sqrt{\Delta_x}(f)=\frac{1}{\Gamma(1/2)}\lim_{y\rightarrow\infty}\int^y\delta(x-y)f(t)(y-t)^{-1/2}dt##

Then the root of delta would be a semi evaluation of f , which you could compute by semi integrating, using Cauchy's formula for repeated integration extended to real values of fractional calculus :

##\sqrt{\Delta_x}(f)=\frac{1}{\Gamma(1/2)}\lim_{y\rightarrow\infty}\int^y\delta(x-y)f(t)(y-t)^{-1/2}dt##

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- #9

anuttarasammyak

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I have never met such a formula before. If it can exist, I suppose

[tex]\int f(x)\sqrt{\delta(x-a)}dx=\int \frac{f(x)}{\sqrt{\delta(x-a)}}\delta(x-a)dx=\frac{f(a)}{\sqrt{\delta(0)}}=0[/tex]

[tex]\int f(x)\sqrt{\delta(x-a)}dx=\int \frac{f(x)}{\sqrt{\delta(x-a)}}\delta(x-a)dx=\frac{f(a)}{\sqrt{\delta(0)}}=0[/tex]

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- #10

wrobel

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and then check that what it is happened does not depend on a sequence

- #11

jasonRF

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EDIT: I just realized that this thread is three years old...

To me it seems like ##\sqrt{\delta(t)}## probably doesn't exist. Here is my thinking (this is not a proof).

First, as suggested by PeroK, start with one particular sequence of finite spikes that converges to a delta function. Let ##p_n(t) = n## for ##-\frac{1}{2n}\leq t\leq\frac{1}{2n}##, and zero everywhere else. If ##\sqrt{\delta(t)}## exists, then I would hope that the sequence ##\sqrt{p_n(t)}## would converge to it. Let ##\phi## be a continuous test function, then

$$

\begin{eqnarray*}

\lim_{n\rightarrow\infty}\int \phi(t) \sqrt{p_n(t)} & = &

\lim_{n\rightarrow\infty} \sqrt{n}\int_{-1/2n}^{1/2n} \phi(t) dt \\

\lim_{n\rightarrow\infty} \sqrt{n}\,\phi(\tau)\int_{-1/2n}^{1/2n} dt

\end{eqnarray*}

$$

for some ##\tau \in [-\frac{1}{2n}, \frac{1}{2n}]##. Then obviously we find that the limit is zero. So we have ##\sqrt{p_n(t)}\rightarrow 0##, even though ##p_n(t)\rightarrow \delta(t)##. This suggests that either ##\sqrt{\delta(t)}=0## or it doesn't exist.

Now I want to look at the product of ##\sqrt{\delta(t)}## with itself. We usually don't talk about multiplying distributions since in general the product of two distributions is not a distribution. However, multiplication of two distributions is allowed in those cases where the product*is* a distribution. In this case, if ##\sqrt{\delta(t)}## is a distribution then obviously it can be multiplied by itself to give the delta distribution. Also, if ##\sqrt{\delta(t)}## is a distribution then it has a Fourier transform that we will denote by ##\mathcal{F}\left[\sqrt{\delta(t)}\right]=F(\omega)##. Now we can use the convolution theorem for Fourier transforms $$

\begin{eqnarray*}

1 & = & \mathcal{F}\left[{ \delta(t)}\right] \

& = & \mathcal{F}\left[\sqrt{ \delta(t)}\sqrt{ \delta(t)}\right] \

& = & \frac{1}{2\pi}F(\omega)\ast F(\omega).

\end{eqnarray*}

$$

I don't know whether a distribution ##F(\omega)## exists that is a constant when convolved with itself. If ##F(\omega)## doesn't exist, then neither does ##\sqrt{\delta(t)}##; if it does exist then clearly ##F(\omega)\neq 0##. However, this seems to contradict our earlier finding that suggests ##\sqrt{\delta(t)}=0## which would imply ##F(\omega)=0##.

So I suspect that ##\sqrt{\delta(t)}## doesn't exist. Or if it does exist, it doesn't mean what the notation suggests it should mean.

jason

To me it seems like ##\sqrt{\delta(t)}## probably doesn't exist. Here is my thinking (this is not a proof).

First, as suggested by PeroK, start with one particular sequence of finite spikes that converges to a delta function. Let ##p_n(t) = n## for ##-\frac{1}{2n}\leq t\leq\frac{1}{2n}##, and zero everywhere else. If ##\sqrt{\delta(t)}## exists, then I would hope that the sequence ##\sqrt{p_n(t)}## would converge to it. Let ##\phi## be a continuous test function, then

$$

\begin{eqnarray*}

\lim_{n\rightarrow\infty}\int \phi(t) \sqrt{p_n(t)} & = &

\lim_{n\rightarrow\infty} \sqrt{n}\int_{-1/2n}^{1/2n} \phi(t) dt \\

\lim_{n\rightarrow\infty} \sqrt{n}\,\phi(\tau)\int_{-1/2n}^{1/2n} dt

\end{eqnarray*}

$$

for some ##\tau \in [-\frac{1}{2n}, \frac{1}{2n}]##. Then obviously we find that the limit is zero. So we have ##\sqrt{p_n(t)}\rightarrow 0##, even though ##p_n(t)\rightarrow \delta(t)##. This suggests that either ##\sqrt{\delta(t)}=0## or it doesn't exist.

Now I want to look at the product of ##\sqrt{\delta(t)}## with itself. We usually don't talk about multiplying distributions since in general the product of two distributions is not a distribution. However, multiplication of two distributions is allowed in those cases where the product

\begin{eqnarray*}

1 & = & \mathcal{F}\left[{ \delta(t)}\right] \

& = & \mathcal{F}\left[\sqrt{ \delta(t)}\sqrt{ \delta(t)}\right] \

& = & \frac{1}{2\pi}F(\omega)\ast F(\omega).

\end{eqnarray*}

$$

I don't know whether a distribution ##F(\omega)## exists that is a constant when convolved with itself. If ##F(\omega)## doesn't exist, then neither does ##\sqrt{\delta(t)}##; if it does exist then clearly ##F(\omega)\neq 0##. However, this seems to contradict our earlier finding that suggests ##\sqrt{\delta(t)}=0## which would imply ##F(\omega)=0##.

So I suspect that ##\sqrt{\delta(t)}## doesn't exist. Or if it does exist, it doesn't mean what the notation suggests it should mean.

jason

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- #12

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- #13

wrobel

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About multiplication of generalized functions see Colombeau theory of generalized functions

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jasonRF

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Do you know of any references on Colombeau theory that are appropriate for non-mathematicians? It looks intriguing, but the references that popped up with a quick google search seem to require a little more mathematical knowledge/sophistication than I have. There are many "simple" treatments of the Schwartz theory of distributions - such as the book by Strichartz (where I first learned it). Is there anything equivalent for the Colombeau theory, or does it inherently require more tools to properly explain?About multiplication of generalized functions see Colombeau theory of generalized functions

thanks,

Jason

- #15

anuttarasammyak

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Supplement to post #9

As eigenfunction of x, ##\delta(x)## have the relation

[tex]f(x) \delta(x) = f(0) \delta(x)[/tex]

If we allow f be not only normal function but also distribution ##\delta^{a}(x)##

[tex]\delta^{a}(x)\delta(x)=\delta^{a}(0)\delta(x)[/tex]

Let a=-1/2

[tex]\delta^{1/2}(x)=\delta^{-1/2}(0)\delta(x)[/tex]

so it is delta function times the coefficient ##\delta^{-1/2}(0)=0##. So

[tex]\delta^{1/2}(x)=0[/tex] in the similar sense with

[tex]x\delta(x)=0[/tex]

when multiplied by normal function f of f(0) ##\neq \infty## and integrated.

As eigenfunction of x, ##\delta(x)## have the relation

[tex]f(x) \delta(x) = f(0) \delta(x)[/tex]

If we allow f be not only normal function but also distribution ##\delta^{a}(x)##

[tex]\delta^{a}(x)\delta(x)=\delta^{a}(0)\delta(x)[/tex]

Let a=-1/2

[tex]\delta^{1/2}(x)=\delta^{-1/2}(0)\delta(x)[/tex]

so it is delta function times the coefficient ##\delta^{-1/2}(0)=0##. So

[tex]\delta^{1/2}(x)=0[/tex] in the similar sense with

[tex]x\delta(x)=0[/tex]

when multiplied by normal function f of f(0) ##\neq \infty## and integrated.

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- #16

wrobel

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Long ago I saw the original Colombeau's monography The new generalized functions and multiplications of distributions. I believe it is appropriate.Do you know of any references on Colombeau theory that are appropriate for non-mathematicians? It looks intriguing, but the references that popped up with a quick google s