In QFT we sometimes encounter functions of the form:

$$K(x-y) = \delta(x-y) + \frac{k}{(x-y)^n} $$

Where $x$ and $y$ are $d$ dimensional vectors and $k$ is a (possibly imaginary) constant. These can arise just from a Fourier transform.

So my question is what happens at $x=y$ ? Is there ever a case where the infinity from the first term cancels out the infinity from the second? (If so what would be the value of $k$, and do they cancel to zero?) Otherwise what are we to make of this expression? If indeed it does cancel to zero at $x=y$ then does it mean the trace of $K$ is zero?


1 Answer 1


Generally, when you have limits that resolve to $\infty \pm \infty$, you have to come up with a strategy to resolve what happens at the singularity, because different ways of approaching infinity will produce different cancellations, and thus, different answers.

In this case, I'd suggest wrapping everything in an integral and seeing how it behaves, since that is the one universal truth of delta functions, and then work backward to determine what happens at $x=y$. In particular, you can only expect the function to be finite there if the indefinite integral is $C^{1}$ continuous there.


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