$Z=\int D[\phi]e^{iS(\phi)}$; is $Z$ real or complex? Is the result of $\int D[\phi]e^{iS(\phi)}$ real or complex?
If it is complex, how does the expectation value for a field, given as the following works?
$$
\langle F\rangle = \frac{\int D[\phi] F(\phi) e^{iS(\phi)}}{\int D[\phi]e^{iS(\phi)}}
$$
If it is real, why is a sum of complex number real - is there an implicit requirement to cancel out the imaginary part?
 A: Let's be extremely careful about the exact path integral formula. We have
\begin{align}
\langle q_f , t_f | {\cal O}(t) |q_i , t_i \rangle &= \int_{q(t_i) = q_i}^{q(t_f)=q_f} [dq(t)] \exp \left[ i \int_{t_i}^{t_f} L ( q(t) , {\dot q}(t) , t ) \right] {\cal O}(t).
\end{align}
We now take a complex conjugate on both sides. On the LHS, we have
$$
\langle q_f , t_f | {\cal O}(t) |q_i , t_i \rangle^* = \langle q_i , t_i | {\cal O}(t)^\dagger |q_f , t_f \rangle
$$
On the RHS, we have
\begin{align}
& \left( \int_{q(t_i) = q_i}^{q(t_f)=q_f} [dq(t)] \exp \left[ i \int_{t_i}^{t_f} L ( q(t) , {\dot q}(t) , t ) \right] {\cal O}(t) \right)^* \\
&\qquad \qquad = \int_{q(t_i) = q_i}^{q(t_f)=q_f} [dq(t)] \exp \left[- i \int_{t_i}^{t_f} L ( q(t) , {\dot q}(t) , t ) \right] {\cal O}(t)^* \\
&\qquad \qquad = \int^{q(t_i) = q_i}_{q(t_f)=q_f} [dq(t)] \exp \left[  i \int_{t_f}^{t_i} L ( q(t) , {\dot q}(t) , t ) \right] {\cal O}(t)^* \\
&= \langle q_i , t_i | {\cal O}(t)^\dagger |q_f , t_f \rangle .
\end{align}
Everything is clearly consistent!
Your case is the same as above with the extension $t_f \to +\infty$, $t_i \to -\infty$.
