directly as the operator $V$ is $\omega$-dependent. However, we can simplify () taking into account that the term $\vert V_{mn}\vert^2$ is symmetric in indices $m,n$ and performing trivial $\alpha$- and $\beta$-integrals. After that the integration over $\omega$ is performed and we get for the electromagnetic current the following expression:

$$\openup 3\jot  \eqalign{
j_{em}^{\mu} &= { N_c\over 4\sqrt{\pi}}
\int_{1/\Lambda^2}^{\infty} { d\tau\over\sqrt{\tau} }
\ \Bigg\lbrace
\sum_n e^{-\tau \epsilon_n^2} \
\langle n \vert {\delta V_0 \over \delta A_{\mu}}
 \, \vert n \rangle \ + \cr
&+ \sum_{m,n}
{e^{-\tau \epsilon_n^2} -e^{-\tau \epsilon_m^2} \over
\epsilon_n^2 - \epsilon_m^2 }
\ \Big[ \
\langle n \vert V_0[A_{\mu}=0] \vert m \rangle
\langle m \vert {\delta V_0 \over \delta A_{\mu} } \vert n \rangle
\ + \cr
&+ {1\over 2\tau} \, \langle n \vert V_1 [A_{\mu}=0]
\vert m \rangle
\langle m \vert {\delta V_1  \over \delta A_{\mu} }
\vert n \rangle \ \Big]
\ \Bigg\rbrace
\ , \cr
}
\eqno(\JFULL)$$

where the notation $V_1 [A_{\mu}=0]$ means the sum of all terms linear in $\omega$ which do not contain $A_{\mu}$. The formula () together with () will be used in the next Section to obtain the electromagnetic form factors’ formulae.

**3. Computing the form factors**

In this point, applying the 