What does it mean to have a divergent path integral in a QFT?
More specifically, if $$\int e^{i S[\phi]/\hbar} D\phi (t)=\infty $$ What does this mean for the QFT of the field $\phi $?
The field $\phi$ has action $$S[\phi]=\int\left(\frac{1}{2}\partial_\mu\phi\partial^\mu\phi-V(\phi)\right)\mbox{d}vol$$ where we use Minkowski signature $(+,-,-,-)$.
If the path integral itself diverges, it means that the v.e.v. diverges. That by itself is bad, because then any arbitrary $n$-point function vanishes. Recall that to compute correlation functions, we append a $J(x)\phi(x)$ to the action and calculate $$ \frac{\delta^n}{\delta J(x_1)\ldots \delta J(x_n)} \int e^{i S[\phi]/\hbar + J(x)\phi(x)}\mathcal{D}\phi = \langle\phi(x_1)\ldots \phi(x_n)\rangle $$ which is normalized by the v.e.v.. Thus, you wouldn't be able to calculate anything sensible.
e.g. a v.e.v. might diverge when upon Wick-rotating to Euclidean time, the action might be unbounded from below (as @Alex points out)- that would typically happen when the potential is bad.
