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Is it possible to calculate the integral: $$J=\int_{-\infty}^{+\infty}f(x)\delta(x-x_0)^kdx$$ wih $k\in\mathbb{R}$? I know that in the Colombeau algebra the distribution $\delta(x)^2$ is defined. What happens if the Delta function is raised to a real number different from $2$? Thanks in advance.

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Yes, $\delta^2$ is defined in the Columbeau algebras, but may result in an infinitesimal number, i.e. a non-zero number smaller than any $1/n$. Colombeau algebras map into a non-standard model of the reals. –  Vobo May 23 '13 at 22:15
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up vote 0 down vote accepted

For $k$ being an integer:

You certainly know that

$\int f(x) \delta (x-x_0) dx = f(x_0)$

This is true regardless of what $f(x)$ is, even when $f(x)$ itself contains a $\delta$-function.

So your integral gives the highly singular result:

$\int f(x)\delta^{k-1}(x-x_0) \delta(x-x_0) dx = f(x_0)\delta^{k-1}(x_0-x_0)$

But you asked about all reals. Don't know what to tell you for non-integer $k$.

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Suppose your answer to be correct. If for instance $k=4$ your result is: $J=f(x_0)\delta^3(x_0-x_0)$ The question now is: what is the meaning of $\delta^3(x_0-x_0)$? –  Riccardo.Alestra May 23 '13 at 16:55
    
I think it is just $\infty$ –  bob.sacamento May 23 '13 at 19:10
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