Dirac delta function confusion

[Antarres]

As a part of a bigger problem, I was trying to evaluate the D'Alambertian of $\epsilon(t)\delta(t^2-x^2-y^2-z^2)$, where $\epsilon(t)$ is a sign function. This term appears in covariant commutator function, so I was checking whether I can prove it solves Klein-Gordon equation. Since there's no mass involved in the term, I would assume that D'Alambertian of the function should be proportional to the function itself(unit mass case).

However after some calculation(product + chain rule), using that $(\epsilon(t))' = 2\delta(t)$ in the sense of distributions, I get after some algebra:
$$\square(\epsilon(t)\delta(\sigma)) = 2\delta'(t)\delta(\sigma) + 8\epsilon(t)\delta'(\sigma)+8t\delta(t)\delta'(\sigma) + 4\epsilon(t)\sigma\delta''(\sigma)$$
where $\sigma = t^2-x^2-y^2-z^2$.
Here, however, I run into problems. For example, the third term in the equation comes from the middle term of product expansion:
$$\square(f(t)g(\sigma)) = \left(\frac{\partial^2}{\partial t^2}f(t)\right)g(\sigma) + 2\frac{\partial}{\partial t}f(t)\frac{\partial}{\partial t}g(\sigma) + f(t)\square g(\sigma)$$
This middle term in our particular case looks like:
$$2\frac{\partial}{\partial t}\epsilon(t)\frac{\partial}{\partial t}\delta(\sigma) =4\delta(t)\frac{\partial \sigma}{\partial t}\frac{d}{d\sigma}\delta(\sigma) = 8t\delta(t)\delta'(\sigma)$$
So the prime means that it's the derivative with respect to $\sigma$, or equivalently, it's just the derivative of the delta function composed to $\sigma$. But then here I run into problems simplifying this, because I feel like I can't employ the normal derivative rule for distributions:
$$\delta'(x)f(x) = -f'(x)\delta(x)$$
I feel like I can pass the derivative this way only for normal partial derivatives(with respect to $t$, $x$, $y$,$z$), but not the $\sigma$ derivatives. So I am unsure how to proceed with simplifying this expression further. It seems to me that if I multiply delta function with some expression, it changes, and that usual rules we have for delta function do not hold if multiplied by some factor.
For example in the factor above I have $t\delta(t)$, which should be evaluated as zero, but I'm unsure if it will be evaluated as zero if it's multiplied by this delta function derivative. And the same goes for the other factors.

I feel like going in circles trying to figure it out, so some clarification would be very helpful.

 

[anuttarasammyak]

Since
$$\delta(t^2-l^2)=\frac{1}{|l|}(\delta(t+l)+\delta(t-l))$$
Except the case x=y=z=0, 8tδ(t)δ′(σ)=0 because points of anomaly, 0, -l, and l with coefficient zero do not coincide.
Then you should focus on the case x=y=z=0. For t=0 8tδ(t)δ′(t^2) is an odd function of t so it should be zero or not defined there sandwitched with divergences of $\pm \infty$ as function $\frac{1}{t}$.

I have a question on the setting of the problem.

As a part of a bigger problem, I was trying to evaluate the D'Alambertian of ϵ(t)δ(t2−x2−y2−z2), where ϵ(t) is a sign function.

Your setting at the origin
$$\epsilon(t)\delta(t^2)$$
seems not clear to me. Fot t=0, this odd function of t is zero or anything else?

 

[jasonRF]

$$\delta(t^2-l^2)=\frac{1}{|l|}(\delta(t+l)+\delta(t-l))$$

I think you mean
$$\delta(t^2-l^2)=\frac{1}{2|l|}(\delta(t+l)+\delta(t-l))$$

jason