Integral over a region in spacetime

Sandra Conor
Hello, can anyone show me if this integral can be evaluated?

##\frac{1}{a_0^2}\int_\Sigma\frac{dy'dz'}{\bigg(y'^2+z'^2+\tfrac{1}{(2a_0)^2}\bigg)^2}##

I am thinking polar coordinates - but what does the summation sign stand for?

Sandra Conor
I have been thinking of polar form too. The summation sign represent a spacelike 2 surface in spacetime.

Initially, I want to evaluate this integral in spacetime.
$$\int_{\Sigma} \frac{dydz}{[a_{o}(y^{2}+z^{2})+2f_{o}y+2g_{o}z+c_{o}]^{2}}$$ where $$a_{o}c_{o}-f_{o}^{2}-g_{0}^{2}=\frac{1}{4}.$$

My way is to define
$$y':=y+\frac{f_0}{a_0},\,z':=z+\frac{g_0}{a_0}$$
Then the quadratic becomes $$a_0(y'^2+z'^2)+c_0-\frac{f_0^2+g_0^2}{a_0}=a_0\bigg(y'^2+z'^2+\tfrac{1}{(2a_0)^2}\bigg).$$

So now, I will need to evaluate$$\frac{1}{a_0^2}\int_\Sigma\frac{dy'dz'}{\bigg(y'^2+z'^2+\tfrac{1}{(2a_0)^2}\bigg)^2}\cdot$$ But this seems complicated. So I am thinking of using polar coordinate and if I do so, it becomes:
$$\frac{1}{a_0^2}\int_\Sigma\frac{dy'dz'}{\bigg(y'^2+z'^2+\tfrac{1}{(2a_0)^2}\bigg)^2} = \frac{1}{a_0^2}\int_\Sigma\frac{rd\phi dr}{\bigg(r^2+\tfrac{1}{(2a_0)^2}\bigg)^2}$$

Since this is in reference to spacetime, I am stuck as to how I should continue. If it is about a disk with radius R then I know what to do but for a spacelike 2 surface, I am still wandering.

• If I haven't done a blunder, I think $\frac{d}{dr}\frac{1}{r^{2}+C}=\frac{-2r}{(r^{2}+C)^{2}}$