There is some additional simplification that can be done which I'll do in a new answer because my browser is not coping well with those large formulas where speed is concerned.
We have $$ \frac{\pi}{2} \sum_{k=0}^{q-1} (-1)^k \cosh(a\rho_k) =
\frac{\pi}{4} \sum_{k=0}^{q-1}(-1)^k e^{a\rho_k} +
\frac{\pi}{4} \sum_{k=0}^{q-1}(-1)^k e^{-a\rho_k}$$
The first sum is
$$ \sum_{k=0}^{q-1} (-1)^k e^{a\rho_k} =
\sum_{k=0}^{q-1} (-1)^k e^{a i \pi /2} e^{a \pi i k} =
e^{a i \pi/2} \frac{1-(-e^{a \pi i})^q}{1 + e^{a \pi i}} $$
which is
$$ e^{a i \pi/2} \frac{1-(-1)^q e^{p \pi i}}{1 + e^{a \pi i}} =
e^{a i \pi/2} \frac{1-(-1)^{p+q}}{1 + e^{a \pi i}} =
e^{a i \pi/2} \frac{2}{1 + e^{a \pi i}} =
\frac{1}{\cos (a\pi/2)}$$
The second sum is
$$ \sum_{k=0}^{q-1} (-1)^k e^{-a\rho_k} =
\sum_{k=0}^{q-1} (-1)^k e^{-a i \pi /2} e^{-a \pi i k} =
e^{-a i \pi/2} \frac{1-(-e^{-a \pi i})^q}{1 + e^{-a \pi i}} $$
which is
$$ e^{-a i \pi/2} \frac{1-(-1)^q e^{-p \pi i}}{1 + e^{-a \pi i}} =
e^{-a i \pi/2} \frac{1-(-1)^{p+q}}{1 + e^{-a \pi i}} =
e^{-a i \pi/2} \frac{2}{1 + e^{-a \pi i}} =
\frac{1}{\cos (a\pi/2)}$$
It follows that the original sum and the integral is
$$J = \frac{\pi}{2} \frac{1}{\cos(a \pi/2)}.$$