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When expanding the scalar field vacuum energy $$\sum_k \frac{1}{2} \omega_k = \frac{1}{2} (L/2\pi)^{n-1} \int \omega(k) d^{n-1}k = \frac{(L^2/4\pi)^{(n-1)/2}}{\Gamma(\frac{n-1}{2})} \int_0^\infty (k^2+m^2)^{\frac {1}{2}}k^{n-1}dk $$ I notice it diverges like $k^n$ for large $k$. I quote the author of the book I'm reading: "the divergence can be usefully analysed by performing the integral with $n$ continued away from integral values to obtain $$ -L^{n-1}2^{-n-1} \pi^{-n/2} m^n\Gamma(-n/2) $$ the gamma function contains poles at all even integrals values for $n\geq0$". It has something to do with dimensional regularization. Can somebody give me some insight on the result of the integral? It is basically a math problem I'm having troubles with. Thank you.

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The substitution $k=m\tan t$ converts the integral to $$\int_0^{\pi/2}m\sec t m^{n-1}\tan^{n-1}t m\sec^2 t dt=m^{n+1}\int_0^{\pi/2}\sin^{n-1}t\cos^{-n-2}t dt.$$We can rewrite this in terms of the Beta function and hence the Gamma function, and overlook the divergence with analytic continuation.

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