Hi Theo,
0) Your definition is roughly correct, yes. For Wiener measure on paths in vector spaces, see Chapter 3 + Appendix A of the 2nd edition of Glimm & Jaffe. On curved targets, I think Bruce Driver has some good lecture notes. One warning: the rough definition of Wiener measure is misleading in one way: Wiener measure is naturally constructed as a measure on a space of distributions which contains the continuous functions.
1) I don't think there's a general theory. Trying a Lagrangian of the form $F = (\dot{\phi})^{1000}$ will not result in happiness. But: You can define Wiener measure using the standard kinetic term $\int \frac{1}{2}|\dot{\phi}|^2dt$ for any $\hbar$, and you can safely add a potential $V$ which is bounded below and integrable to the kinetic term.
2) Any observable you can write down should give you a Wiener integrable function, in quantum mechanics. This is not true in QFT, however. Most of the work in constructive QFT is proving that the measure on the space of histories actually has moments!
Consider for any real number $s>1$ the double series
$$
\sum_{m=1}^{\infty} \sum_{k=0}^{\infty} \frac{(-1)^m}{((2k+1)^2+m^2)^s}. \tag{1}
$$
This double series converges absolutely. Moreover for any given $m$, the inner series over $k$ converges to some $C(m,s)$, say, which for any given $s \ge 1$ is a monotone decreasing function of $m$. Combined with (the idea behind the) alternating series test, it is not hard to justify that the series in (1) tends to our desired sum as $s\to 1^+$.
Now let $R(n)$ denote the number of ways of writing $n$ as $a^2+b^2$ with $a$ and $b$ integers; note $R(1)=4$ since $1=1^2+0^2= (-1)^2+0^2=0^2+1^2=0^2+(-1)^2$. It is well known that
$$
\sum_{n=1}^{\infty} \frac{R(n)}{n^s} = 4\zeta(s) L(s,\chi_{-4}),
$$
and that
$$
\sum_{n=1, \text{n odd}}^{\infty} \frac{R(n)}{n^s} = 4 \Big(1-\frac{1}{2^s}\Big) \zeta(s)L(s,\chi_{-4}).
$$
Here $L(s,\chi_{-4})=1/1^s-1/3^s+1/5^s-1/7^s+\ldots$ is the Dirichlet $L$-function for the character $\pmod 4$.
Now let us return to the sum in (1). Write $(2k+1)^2+m^2=n$. If $n$ is odd, then $m$ is necessarily even, and this number is counted in (1) a total of $R(n)/8$ times; the only exceptions are when $n$ is an odd square, where the solutions $(2k+1)^2+0^2$ are not counted. So the contribution of the odd numbers $n$ to (1) is
$$
\frac{1}{8} \sum_{n \text{odd} } \frac{R(n)}{n^s} - \frac{1}{2} \sum_{k=0}^{\infty} \frac{1}{(2k+1)^{2s}}= \frac{1}{2} \Big(1-\frac{1}{2^s}\Big) \zeta(s)L(s,\chi_{-4}) - \frac 12 \Big(1-\frac{1}{2^{2s}}\Big) \zeta(2s).
$$
Now consider the contribution of $n\equiv 2\pmod 4$ to (1). These are the terms with $m$ odd, and they appear with sign $-1$. Here each $n$ appears $R(n)/4$ times. So these terms give
$$
-\frac{1}{4} \sum_{n\equiv 2\pmod 4} \frac{R(n)}{n^s} = -\frac{1}{4} \frac{1}{2^s} \sum_{\ell \text{ odd} }\frac{R(\ell)}{\ell^s} = -\frac{1}{2^{s}}\Big(1-\frac{1}{2^s}\Big)\zeta(s) L(s,\chi_{-4}),
$$
upon writing $n=2\ell$ with $\ell$ odd, and using here $R(n)=R(\ell)$.
Thus our sum in (1) equals
$$
\Big(\frac12 -\frac{1}{2^s}\Big) \Big(1-\frac{1}{2^s}\Big) \zeta(s) L(s,\chi_{-4}) - \frac 12 \Big(1-\frac{1}{2^{2s}}\Big) \zeta(2s).
$$
Now let $s\to 1^+$ and use $\zeta(2)=\pi^2/6$, $L(1,\chi_{-4})=\pi/4$ (Gregory's formula), and $(1/2-1/2^s) \zeta(s) \to (\log 2)/2$ (zeta has a simple pole with residue $1$ at $1$). The claimed identity follows.
Best Answer
Using Christian Krattenthaler's hyp.m tells you the following - the important information is at the end, and the result agrees with Johannes' comment.