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When calculating integrals with delta functions, the integration domains are usually rectangular. However, recently, I encountered a set of time ordered integrals over Dirac deltas of the following form: $$ \int_0^1 dt_1 \int_0^{t_1} dt_2 \int_0^{t_2} dt_3 \int_0^{t_3} dt_4 K(t_1,t_2,t_3,t_4) \delta(t_1-t_2)\delta(t_3-t_4) $$ $$ \int_0^1 dt_1 \int_0^{t_1} dt_2 \int_0^{t_2} dt_3 \int_0^{t_3} dt_4 K(t_1,t_2,t_3,t_4) \delta(t_1-t_3)\delta(t_2-t_4) $$ $$ \int_0^1 dt_1 \int_0^{t_1} dt_2 \int_0^{t_2} dt_3 \int_0^{t_3} dt_4 K(t_1,t_2,t_3,t_4) \delta(t_1-t_4)\delta(t_3-t_2) $$ Where $K(t_1,t_2,t_3,t_4)$ is some smooth function of the arguments.

For convenience, I set $K=1$ and evaluated these integrals in mathematica. All three of them are zero! This answer makes no sense to me, because suppose I add up these three integrals (for $K=1$), I would get: $$ \int_0^1 dt_1 \int_0^{t_1} dt_2 \int_0^{t_2} dt_3 \int_0^{t_3} dt_4 \bigg(\delta(t_1-t_2)\delta(t_3-t_4)+\delta(t_1-t_3)\delta(t_2-t_4)+\delta(t_1-t_4)\delta(t_2-t_3)\bigg) $$ This expression is manifestly symmetric wrt to all permutation of $t_i$'s. Therefore, the time ordered integral can be transformed into a regular integral over rectangular domains: $$ \frac{1}{24} \int_0^1 dt_1 \int_0^{1} dt_2 \int_0^{1} dt_3 \int_0^{1} dt_4 \bigg(\delta(t_1-t_2)\delta(t_3-t_4)+\delta(t_1-t_3)\delta(t_2-t_4)+\delta(t_1-t_4)\delta(t_2-t_3)\bigg) $$ This integral evaluates to $\frac{1}{8}$ instead of zero.

So the question is: which one is the correct answer? And more generally, how should I think about Dirac deltas rigorously in time-ordered integrals?

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    $\begingroup$ How does one define the functional $\int_0^\infty \delta(x)\,dx$? It isn't uniquely defined. $\endgroup$ – Mark Viola Jul 4 '18 at 20:27
  • $\begingroup$ The integral as stated is not defined (as Mark Viola has pointed out). You have to go back a few steps in your calculation to understand how the expression came about and at which step the problem appeared. You integral looks like it might be a physics problem and there sometimes stuff like causality comes to rescue... $\endgroup$ – Fabian Jul 4 '18 at 21:59
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  • We can replace simplex integration region with a rectangular domain $[0,1]^4$ if we multiply the integrand with $$ \theta(t_1\!-\!t_2)\theta(t_2\!-\!t_3)\theta(t_3\!-\!t_4), \tag{A}$$ where $\theta$ denotes the Heaviside step function. Multiplying distributions with discontinuous functions is not mathematically well-defined.

  • Nevertheless, we cannot resist to heuristically calculate what would happen with OP's integral in the case $K=1$ if we assume that $\theta(0)=\frac{1}{2}$.

    1. If we first integrate over $t_1$ & $t_4$ the first integrand becomes $$\theta(0)\theta(t_2\!-\!t_3)\theta(0).$$ The integrations over $t_2$ & $t_3$ then yield $\frac{1}{8}$.

    2. If we first integrate over $t_3$ & $t_4$ the second integrand becomes $$\theta(t_1\!-\!t_2)^2\theta(t_2\!-\!t_1) ~=~\frac{1}{8}\delta_{t_1,t_2}~=~0\text{ a.e.}$$ Here $\delta_{t_1,t_2}$ is the Kronecker delta function$^1$, which is zero almost everywhere (a.e.). The integrations over $t_1$ & $t_2$ then yield $0$.

    3. Similarly, if we first integrate over $t_3$ & $t_4$ the third integrand becomes $$\theta(t_1\!-\!t_2)\theta(0)\theta(t_2\!-\!t_1) ~=~\frac{1}{8}\delta_{t_1,t_2}~=~0\text{ a.e.}$$ The integrations over $t_1$ & $t_2$ then yield $0$.

    Hence the total integral is heuristically $\frac{1}{8}+0+0=\frac{1}{8}$, which agrees with OP's symmetrization argument.

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$^1$The Kronecker delta function $\delta_{t_1,t_2}$ should not be confused with the Dirac delta distribution $\delta(t_1\!-\!t_2)$.

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  • $\begingroup$ +1. I'll add that this has some physical content, it corresponds to passing to a symmetric approximate identity rather than an arbitrary approximate identity. As mentioned in physics.stackexchange.com/a/61338/108628 (an answer you wrote to the question you linked), this is closely related to the Cauchy principal value, it is not just a superficial resemblance. $\endgroup$ – Ian Jul 6 '18 at 14:19
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One trick:

Whenever you have $\delta (t_1 - t_2) \delta (t_3 - t_4)K(t_1, t_2, t_3, t_4)$ you can replace the integrals to be over just two variables, $t_1$ and $t_3$, say, and argument $K(t_1, t_1, t_3, t_3)$.

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  • $\begingroup$ Hey David, that's exactly what I thought would be true. But for some reason, mathematica didn't agree with that, which is why I am confused. $\endgroup$ – Zhengyan Shi Jul 4 '18 at 19:03
  • $\begingroup$ I suspect this is a problem with the algorithmics in Mathematica rather than a misunderstanding of actual mathematics. $\endgroup$ – David G. Stork Jul 4 '18 at 20:01

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