C. Explicit computation of the vacuum energy integral

We concentrate in this appendix on the computation of the contribution $I ( k , \bar{k} ; \kappa , m)$ of equation (28). Of course since the integral is divergent, it must be properly regularized and we choose to do that by putting an infrared ($\epsilon$) and an ultraviolet ($\Lambda$) cutoff on the modulus of the momentum $k$ and of its projection $\bar{k}$, exploiting some of the arbitrariness in the choice of the regularization scheme. Thus integrals written with implicit integration domain, like $\int d^{4} k (\dots)$, are to be understood as performed in the domain of the variables $( k _{0}, k _{1}, k _{2}, k _{3} )$, which is the inverse image of the domain $\epsilon \leq r \leq \Lambda$, $0 \leq \vartheta < 2 \pi$, $\epsilon \leq \rho \leq \Lambda$, $0 \leq \varpi < 2 \pi$ in the variables $(r , \rho , \vartheta , \varpi)$ under the following change of variables in euclidean space:

$$\pmatrix{ k ^{4} \cr k ^{1} \cr k ^{2} \cr k ^{3}} = T ( r ,... ... r \sin \vartheta \cr \rho \cos \varpi \cr \rho \sin \varpi }$$

with jacobean

$${\mathcal{J}} T = \pmatrix{ \cos \vartheta & - r \sin \varth... ...ho \sin \varpi \cr 0 & 0 & \sin \varpi & \rho \cos \varpi } ,$$

whose determinant is

$$\det \left( {\mathcal{J}} T \right) = r \rho$$

so that

$$d ^{4} k = r \rho d r d \rho d \vartheta d \varpi \,.$$

We thus have

$$I ( k , \bar{k} ; \kappa , m _{\mathrm{A}} ) =$$

$$\quad = \int \frac{d^4k}{(2\pi)^4} \ln \left[ k ^{2} ( k ^{2} - m _{\mathrm{A}} ^{2}) - \kappa \bar{k} ^{2} \right]$$

$$\quad = \frac{1}{(2 \pi) ^{4}} \int _{0} ^{2 \pi} \! \! \! \! \! ... ... ^{2} ) ( r ^{2} + \rho ^{2} - m _{\mathrm{A}} ^{2}) - \kappa \rho ^{2} \right]$$

$$\quad = \frac{1}{4 \pi ^{2}} \int _{\epsilon} ^{\Lambda} \! \! \!... ...{A}} ) r ^{2} + \rho ^{2} ( \rho ^{2} - m ^{2} _{\mathrm{A}} - \kappa ) \right]$$

$$\quad \equiv \frac{1}{4 \pi ^{2}} \left . J ( r , \rho ; \kappa , m _{\mathrm{A}}) \right \vert _{r , \rho = \epsilon} ^{r , \rho = \Lambda}\,.$$ (37)

We have thus the problem of computing $J ( r , \rho ; \kappa , m _{\mathrm{A}})$, which can be calculated in closed form: the final result can be expressed as

$$J ( r , \rho ; \kappa , m _{\mathrm{A}} ) = \frac { \rho ^{2} \left[ \kappa + 3 ( m _{\mathrm{A}} ^{2} - 6 r ^{2} - \rho ^{2} ) \right] } {24} +$$

$$+ \, \frac{\mbox{\boldmath$B$}^{3/2}}{48\kappa} \ln \left( \frac{... ...dmath$A$}}}{\kappa + \mbox{\boldmath$C$}+ \sqrt{\mbox{\boldmath$A$}}} \right) +$$

$$+\, \frac { \left[ \kappa ^{2} + 3 \kappa (m _{\mathrm{A}} ^{2} -... ...$D$}) \right] } {48} \ln \left( - \kappa \rho ^{2} + \mbox{\boldmath$D$}\right)$$ (38)

if we set

$$\mbox{\boldmath$A$} = A (r ; m _{\mathrm{A}} , \kappa) \equiv ( \kappa + m _{\mathrm{A}} ^{2} ) ^{2} - 4 \kappa r ^{2}$$

$$\mbox{\boldmath$B$} = B (\rho ; m _{\mathrm{A}} , \kappa) \equiv 4 \kappa \rho ^{2} + m _{\mathrm{A}} ^{4}$$

$$\mbox{\boldmath$C$} = C (r , \rho ; m _{\mathrm{A}}) \equiv m _{\mathrm{A}} ^{2} - 2 ( r ^{2} + \rho ^{2} )$$

$$\mbox{\boldmath$D$} = D (r, \rho ; m _{\mathrm{A}}) \equiv ( r ^{2} + \rho ^{2} ) ^{2} ... ... \frac{\mbox{\boldmath$C$}^{2} - \mbox{\boldmath$B$}}{4} + \kappa \rho ^{2} \,.$$

The final stage consists in evaluating it in the infrared ( $\epsilon \to 0$) and ultraviolet limits. In this case this is equivalent to the computation of

$$\lim _{(r , \rho) \to (0 , 0)} J ( r , \rho ; \kappa , m _{\mathrm{A}} )$$

and the extraction of the divergent contribution in

$$J ( r = \Lambda , \rho = \Lambda ; \kappa , m _{\mathrm{A}} )$$

when $\Lambda \to \infty$. This is done in the next subsections.