Most probably this question should be well studied in the theory of stochastic processes, but I am not educated in that area. Sorry if this question is too elementary.


Let $V\colon \mathbb{R}\to \mathbb{R}$ be a function to be specified later. Fix two points $a,b\in \mathbb{R}$. I am interested in convergence of the following path integral 
$$I:=\int \mathcal{D}x \exp\left\{-\int\_0^L\left(\frac{1}{2}\cdot\left(\frac{dx(\tau)}{d\tau}\right)^2 +V(x(\tau))\right)d\tau\right\},$$
where the integral is over all maps $x\colon [0,L]\to \mathbb{R}$ such that $x(0)=a,x(L)=b$.


If I am not mistaken, this integral $I$ can be rewritten as integral with respect to the Wiener measure on the space of continuous functions.



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> Under what conditions on $V$ one can prove convergence of $I$ mathematically rigorously? For example, $V$ being a polynomial of fourth degree with positive leading coefficient?
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Convergence of an integral with respect to the Wiener measure
The conditional Wiener measure is concentrated on the space $C(L,a,b)$ of continuous curves $x : [0,L] \to \mathbb R$ such that $x(0) = a$ and $x(L) = b$, endowed with the topology given by the distance $d(x,y)= \sup \_{t \in [0,L]} |x(t) - y(t)|$. Since it is a Borel, regular measure, all continuous and bounded functions on $C(L,a,b)$ will be integrable with respect to it.


Notice now that if $\inf V > -\infty$, then the function $\widetilde V (x) = \exp (- \int \_0 ^L V(x(t)) \ \mathrm d t)$ is lower-bounded by $0$ and upper-bounded by $\exp (-L \inf V)$, therefore it is bounded.


Assuming $V$ continuous let us show that $\widetilde V$ is continuous, too. Let $r>0$. If $d(x\_n, x) \to 0$ then there exists $n\_r$ such that for $n \ge n\_r$ we shall have $d(x\_n, x) < r$, which implies that for $t \in [0,L]$


$$|x\_n(t) - a| = |x\_n (t) - x(t) + x(t) - x(a)| \le |x\_n (t) - x(t)| + |x(t) - x(a)| \le \\
\le \sup \_{t \in [0,L]} |x\_n (t) - x(t)| + \sup \_{t \in [0,L]} |x (t) - x(a)| = r + \sup \_{t \in [0,L]} |x (t) - x(a)| =R$$


which shows that for sufficiently large $n$ the curves $x\_n$ will live inside the ball


$$B(a, R) = \{u \in \mathbb R \mid |u-a| \le R\}$$


which is compact. Since $V$ is continuous, its restriction $V \big| \_{B(a,R)}$ will be bounded by some $M \ge 0$, therefore $V \circ x\_n$ will be bounded by $M$ for $n \ge n\_r$, and therefore (by possibly increasing $M$) for all $n$ (because there are only finitely many $n < n\_r$). Also, if $d(x\_n, x) \to 0$ it follows that $V \circ x\_n \to V \circ x$. We are therefore within the hypotheses of Lebesgue's dominated convergence theorem and we may conclude that $\widetilde V (x\_n) \to \widetilde V (x)$, which shows that $\widetilde V$ is continuous.


To conclude, if $V$ is continuous and bounded from below then $\widetilde V$ will be continuous and bounded, therefore integrable according to the first paragraph, i.e. the integral that you are investigating will converge.


The above are only sufficient conditions, not necessary ones. In particular, one may relax continuity and work in an only measurable context, but for most concrete applications the above should suffice.


