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I have come across the expression $$ \int f(x) \delta(x-a) \delta''(x-a) \mathrm dx$$ where the prime represents the derivative.

Usually with derivatives of the Dirac delta distribution I'd partially integrate, but here I keep running in circles. What I tried is \begin{align} \int f \delta \delta'' &= - \int (f \delta)' \delta' = -\int f' \delta \delta' - \int f \delta' \delta' = - \int f' \delta \delta' + \int (f \delta')' \delta\\ & ={} - \int f' \delta \delta' + \int f' \delta' \delta + \int f \delta '' \delta= \int f \delta \delta'' \end{align} or if I take the other term for the second partial integration \begin{align} \int f \delta \delta'' &= - \int (f \delta)' \delta' = \int (f \delta)'' \delta = \int f'' \delta \delta + 2 \int f' \delta' \delta + \int f \delta '' \delta \\ \Rightarrow 0 &= \int f'' \delta \delta + 2 \int f' \delta' \delta \end{align} which I also could have gotten from elementary partial integration of $ f'' \delta^2 $.

What further options do I have? Squares of delta functions etc. are not a problem.

Qmechanic
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Neuneck
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1 Answers1

2

I was just playing with things and got this, let me know if you find it a useful idea (or if I'm missing something which makes this useless).

Denote $\delta_k(x)=\frac{1}{k\sqrt{\pi}} e^{-(x/ k)^2}$. You get the well known (completely mathematically rigorous) result:

$\lim_{k\to 0} \int_{-\infty}^\infty f(x) \delta_k(x) dx=f(0)$ for well behaved enough $f$.

So let's look at:

$\lim_{k\to 0} \int_{-\infty}^\infty (c_0+c_1 x+c_2 x^2+\cdots) \delta_k''(x) \delta_k(x) dx=f(0)$

Using the well known technique of Proof By Mathematica:

delta[x_] := 1/(k Sqrt[ Pi]) E^(-(x/k)^2);
g[x_] = FullSimplify[delta''[x] delta[x] x^n];
Integrate[g[x], {x, -Infinity, Infinity}, Assumptions -> {Element[n, Integers], a > 0}]

I get the result:

$$\int_{-\infty}^\infty x^n \delta_k''(x) \delta_k(x) dx=\frac{2^{-\frac{n}{2}-\frac{3}{2}} \left((-1)^n+1\right) (n-1) k^{n-3} \Gamma \left(\frac{n+1}{2}\right)}{\pi }$$

As $k\to 0^+$, the value is zero for $n=1,3,4,5,6,7,\cdots$ and diverges for $n=0,2$, like $-\frac{1}{\sqrt{2 \pi}k^3}$ for the $n=0$ case and like $\frac{1}{4 k\sqrt{2 \pi}}$ for the $n=2$ case.