The identity

$$x^p \; \delta^{(n)}(x) = (-1)^p \frac{n!}{(n-p)!} \; \delta^{(n-p)}(x)$$

can easily be derived from the generalized Leibnitz formula for $n$ and $p$ positive integers:

$$\int \; x^p \;\delta^{(n)}(x)\; g(x) \; dx= (-1)^nD_{x=0}^n\; x^p g(x)=(-1)^n \sum_{k=0}^n \binom{n}{k}D_{x=0}^k \; x^p D^{n-k}_{x=0} \; g(x).$$

An alternative, simpler symbolic derivation employs another rep of the Dirac delta using the Heaviside step function (cf. Fractional calculus and interpolation of generalized binomial coefficients):

$$x^p \; \delta^{(n)}(x) = x^p H(x) \frac{x^{-n-1}}{(-n-1)!}= \frac{(p-n-1)!}{(-n-1)!} H(x) \frac{x^{p-n-1}}{(p-n-1)!}$$

$$=(-1)^p \frac{n!}{(n-p)!} \; \delta^{(n-p)}(x)$$

noting that $\binom{s-1+p}{p}=(-1)^p \binom{-s}{p}$.

These results are, of course, consistent with the Fourier transform rep:

$$\int^{\infty}_{-\infty}\; x^p \; \delta^{(n)}(x) \; \exp(-i2 \pi fx) \; dx = \frac{1}{(-i2 \pi)^p}D^p_f\int^{\infty}_{-\infty}\; \delta^{(n)}(x) \; \exp(-i2 \pi fx) \; dx $$

$$=\frac{1}{(-i2 \pi)^p}D^p_f \; (i2 \pi f)^n=\int^{\infty}_{-\infty}\; (-1)^p \frac{n!}{(n-p)!} \; \delta^{(n-p)}(x) \; \exp(-i2 \pi fx) \; dx.$$

These last two methods give sensible consistent results for $p$ extended to negative integers and $$D_f^{p} w(f) =H(f)\int_0^f \frac{(f-t)^{-p-1}}{(-p-1)!} w(t) \; dt$$ analytically continued from positive to negative values of $f$ and $w(f)$ analytic about $f=0$, i.e., with $D_f^p$ as the repeated derivative or integration op (antiderivative).

So, in this sense you could claim that the top identity holds for $p$ extended to any integer, not just for positive integers.

Has anyone any additional arguments for this claim or a counter-argument other than the usual mantra that products of distributions are typically ill-defined?

(This has some bearing on the representation of integration in Fourier space as a division.)

  • $\begingroup$ What do you understand under $(-n-1)!$? The gamma function is underfined for negative integers $\endgroup$ – TZakrevskiy Aug 14 '15 at 9:30
  • $\begingroup$ @TZakrevskly, the reciprocal gamma function with complex argument is an entire function with zeros at the neg integers. See tcjpn.wordpress.com/2015/08/15/… $\endgroup$ – Tom Copeland Aug 16 '15 at 15:50
  • $\begingroup$ See also pg. 26 of "Multiple polylogarithms and mixed Tate motives" by Goncharov arxiv.org/abs/math/0103059. $\endgroup$ – Tom Copeland Jun 26 '17 at 16:09

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