Summing over zero mode energies

I am reading Birrell's and Davies' book on QFT in Curved Spacetime. In Chapter two, the authors try to analyze the divergence associated with the infinite constant, originating from the sum over all the zero oscillatory modes. I.e., $$\frac{1}{2}\sum_{k}\omega_k=\Big(\frac{L^2}{4\pi}\Big)^{(n-1)/2} \frac{1}{\Gamma(\frac{n-1}{2})}\int_0^{\infty}dk \ k^{n-2}(k^2+m^2)^{1/2}\;,$$ where $$k$$ is the magnitude of the vector $$\vec{k}$$, $$\omega_k=\sqrt{k^2+m^2}$$ is the energy associated with the mode carrying that momentum, and the $$\Gamma$$ function originates from calculating the area of an $$n-1$$ dimensional unit sphere.

The result is something of the form: $$-L^{n-1}2^{-n-1}\pi^{-n/2} m^n \Gamma\big(-\frac{n}{2}\big)$$ The authors claim that the integral is performed by "continuing $$n$$ away from the integral values". What exactly does that phrase mean? How exactly do they derive the second expression from the first?

The only thing I can explain (despite not being sure about it) is the $$m^n$$ factor, by assuming that at some point they make a substitution of a sort $$k=mx$$ and the integral reduces to an integral over $$x$$.

Any help will be appreciated.

The authors claim that the integral is performed by "continuing n away from the integral values". What exactly does that phrase mean?

It means that they stop treating the symbol "n" as representing an integer (e.g., 1, 2, 3, etc) and start treating that symbol as representing a complex number. This is sometimes called "analytic continuation."

The procedure assumes that there is an analytic (in some region) function of a complex variable "z" that takes on the correct values when z is an integer: $$I(z)=\Big(\frac{L^2}{4\pi}\Big)^{(z-1)/2} \frac{1}{\Gamma(\frac{z-1}{2})}\int_0^{\infty}dk \ k^{z-2}(k^2+m^2)^{1/2}\;.$$

How exactly do they derive the second expression from the first?

Compare the integral to the definition of the Gamma function: $$\Gamma(z) = \int_0^\infty x^{z-1}e^{-x} dx\;.$$

The above definition does not immediately give you the integral, but hopefully points you in the right direction. You will have to make some substitutions/changes of variable. You might also have to deal with the branch points related to the square root, but I don't think it's going to be an issue here. You might also want to compare the Gamma function or products of the Gamma function to other integral representations such as the ones in the section "Relation to other functions" in the wikipedia article on the gamma function like: $$B(x,y) = \int_0^1 t^{x-1}(1-t)^{y-1}dt = \frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}\;.$$

• Then one last formula makes sense of $\Gamma(-n/2)$, which still diverges for even $n\ge0$.
– J.G.
Commented Jul 7, 2022 at 17:14