C..1 Infrared limit

=1 In this subsection we consider the infrared limit and in the following we will use the symbol ``$\cong$'' to imply that two expression are equivalent in the infrared limit, i.e. they have the same limit. Moreover we will get rid of the square roots by means of the following results

$$\sqrt{ 1 - \frac{4 \kappa r ^{2}}{( \kappa + m _{\mathrm{A}} ^{2} ) ^{2}} } \cong 1 - \frac{2 \kappa r ^{2}}{\vert \kappa + m _{\mathrm{A}} ^{2} \vert ^{2}}$$ (39)

$$\sqrt{1 + \frac{4 \kappa \rho ^{2}}{m _{\mathrm{A}} ^{4}}} \cong 1 + \frac{2 \kappa \rho ^{2}}{m _{\mathrm{A}} ^{4}} .$$ (40)

We now turn to the contributions in the various lines of equation (38). The first one gives no problem:

$$\frac { \rho ^{2} \left[ \kappa + 3 ( m _{\mathrm{A}} ^{2} - 6 r ^{2} - \rho ^{2} ) \right] } {24} \cong 0 \,.$$ (41)

The second one has a well behavior in the $\mbox{\boldmath$B$}$ term; more care has to be paid in the logarithm:

$$\frac{\mbox{\boldmath$B$}^{3/2}}{48\kappa} \ln \left( \frac{\mbox... ...{\mbox{\boldmath$B$}}}{\mbox{\boldmath$C$}+ \sqrt{\mbox{\boldmath$B$}}} \right) \cong \frac{m _{\mathrm{A}} ^{6}}{48 \kappa} \ln \left[ \frac{ m _{\mat... ...^{2} + \rho ^{2} ) + \sqrt{4 \kappa \rho ^{2} + m _{\mathrm{A}} ^{4}} } \right]$$

$$\cong \frac{m _{\mathrm{A}} ^{6}}{48 \kappa} \ln \left[ \frac{ m _{\mat... ...2} \left( 1 + \frac{2 \kappa \rho ^{2}}{m _{\mathrm{A}} ^{4}} \right) } \right]$$

$$\cong \frac{m _{\mathrm{A}} ^{6}}{48 \kappa} \ln \left[ - m _{\mathrm{A... ...A}} ^{6}}{48 \kappa} \ln \left[ {( m _{\mathrm{A}} ^{2} ) ^{2}} \right] .\qquad$$ (42)

The last term on the second line of (38) again gives troubles only inside the logarithmic term, which can be elaborated as follows

$$\ln \left( \frac{\kappa + \mbox{\boldmath$C$}- \sqrt{\mbox{\boldmath$A$}}}{\kappa + \mbox{\boldmath$C$}+ \sqrt{\mbox{\boldmath$A$}}} \right) = \ln \left[ \frac{ \kappa + m _{\mathrm{A}} ^{2} - 2 ( r ^{2} + \r... ...} ) + \sqrt{( \kappa + m _{\mathrm{A}} ^{2} ) ^{2} - 4 \kappa r ^{2}} } \right]$$

$$\cong \ln \left[ \frac{ \kappa + m _{\mathrm{A}} ^{2} - 2 ( r ^{2} + \r... ...appa r ^{2}} {\vert \kappa + m _{\mathrm{A}} ^{2} \vert ^{2}} \right) } \right]$$

$$\cong \cases{ \ln \left[ \frac{ - ( r ^{2} + \rho ^{2} ) + \frac{\kappa... ...} } \right) ^{-1} \right] \quad \mbox{if $\kappa + m _{\mathrm{A}} ^{2} < 0$} }$$

$$\cong \ln \left[ \left( \frac{ - ( r ^{2} + \rho ^{2} ) + \frac{\kappa ... ...{A}} ^{2}} } \right) ^{\mathrm{Sign} ( \kappa + m _{\mathrm{A}} ^{2} )} \right]$$

$$\cong \mathrm{Sign} ( \kappa + m _{\mathrm{A}} ^{2} ) \! \left\{ \ln \!... ...ln \! \! \left[ ( \kappa + m _{\mathrm{A}} ^{2} ) ^{2} \right] \right\} .\qquad$$ (43)

Since we also have

$$\frac{\mbox{\boldmath$A$}^{3/2}}{48 \kappa} \cong \frac{\vert \kappa + m ^{2} _{\mathrm{A}} \vert ^{3}}{48 \kappa} \,,$$ (44)

the two previous results, (44) and (43), combine in a neat way: the sign in the first factor exactly combines with the absolute value of the second factor

$$\vert \kappa + m _{\mathrm{A}} ^{2} \vert ^{3} \mathrm{Sign} ( \kappa + m _{\mathrm{A}} ^{2} ) \,,$$

so that

$$\frac{\mbox{\boldmath$A$}^{3/2}}{48 \kappa} \ln \left( \frac{\kap... ...oldmath$A$}}}{\kappa + \mbox{\boldmath$C$}+ \sqrt{\mbox{\boldmath$A$}}} \right) \cong \frac{( \kappa + m _{\mathrm{A}} ^{2} ) ^{3}}{48 \kappa} \ln \left[ - m _{\mathrm{A}} ^{2} ( r ^{2} + \rho ^{2} ) - \kappa \rho ^{2} \right]-$$

$$- \, \frac{( \kappa + m _{\mathrm{A}} ^{2} ) ^{3}}{48 \kappa} \ln \left[ ( \kappa + m _{\mathrm{A}} ^{2} ) ^{2} \right] .$$ (45)

For the last term in (38) we do not have too much work. The factor before the logarithm has no problems and we can simply forget about the $r$ and $\rho$ dependent parts. Instead inside the logarithm we can neglect higher order terms in the limit we are interested in, so that

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

$$\qquad \cong \frac{ \kappa ^{2} + 3 \kappa m _{\mathrm{A}} ^{2} +... ...eft[ - m _{\mathrm{A}} ^{2} ( r ^{2} + \rho ^{2} ) - \kappa \rho ^{2} \right] .$$ (46)

The desired result, $J ( 0 , 0 ; \kappa , m _{\mathrm{A}})$ is then (41) $+$ (42) $-$ (45) $+$ (46), i.e.

$$J ( 0 , 0 ; \kappa , m _{\mathrm{A}}) = \frac{m _{\mathrm{A}} ^{6}}{48 \kappa} \left \{ \ln \left[ - m _{... ...o ^{2} \right] - \ln \left[ {( m _{\mathrm{A}} ^{2} ) ^{2}} \right] \right \} +$$

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

$$+\, \frac{( \kappa + m _{\mathrm{A}} ^{2} ) ^{3}}{48 \kappa} \ln \left[ ( \kappa + m _{\mathrm{A}} ^{2} ) ^{2} \right] +$$

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

$$= \frac{( \kappa + m _{\mathrm{A}} ^{2} ) ^{3}}{24 \kappa} \ln \lef... ..._{\mathrm{A}} ^{2} ) ^{3}}{24 \kappa} \ln \left( m _{\mathrm{A}} ^{2} \right) .$$ (47)

As also pointed out in the main text this contribution is finite in the case of vanishing external field ($\kappa \to 0$), since

$$\lim _{\kappa \to 0} \left [ \frac{( \kappa + m _{\mathrm{A... ...athrm{A}}}{24} \left[ \ln (m ^{6} _{\mathrm{A}}) + 1 \right].$$ (48)