Integral with Cutoff: Understand the Logarithmic Terms

[LAHLH]

Hi,

I'm struggling with how to see that

\int^{L}_{0}\int^{L}_{0} \frac{dxdy}{(x-y)^2}=2L/a-2\ln{(L/a)}+\mathcal{O}(1)

'a' here is a cutoff to avoid the divergence that occurs when x=y, I presume we just set the integrand to zero when x-y<a, I think.

Can anyone see why the above holds?

 

[HallsofIvy]

What you written makes no sense. If there was no "a" in the integral, there cannot be an "a" in the result.

 

[LAHLH]

Well it's something the author does 'by hand' I guess, i.e. if x-y<a then set the integrand to zero, type approach. It's from Srednicki ch82 for some context and 'a' is lattice spacing. Since obviously the left hand side of the above integral is divergent really.

Based on some similar integrals I've seen in this subject I was thinking the method could work something like: integrand only depends on difference (x-y) thus might as well fix y at a given point say 0, then the y integral just contributes L and you get L \int^{L}_0 1/(x-0)^2, but now set the integrand to zero when x-y<a => L \int^{L}_a 1/(x)^2 =L[-1/L+1/a]=L/a-1

which is sort of getting there but not quite..