What proof of quadratic reciprocity is Hilbert referring to in this quote? Let $(a, b)_v$ denote the Hilbert symbol on the completion $K_v$ of a global field $K$ at a place $v$.  The Hilbert reciprocity law $\prod_v (a, b)_v = 1$ is a strict generalization of quadratic reciprocity, to which it reduces in the case $K = \mathbb{Q}, a = p, b = q$.  Hilbert had this to say about his law:
The reciprocity law... reminds [sic] the Cauchy integral theorem, according to which the integral of a function over a path enclosing all of its singularities always yields the value $0$.  One of the known proofs of the ordinary quadratic reciprocity law suggests an intrinsic connection between this number-theoretic law and Cauchy's fundamental function-theoretic theorem.
(I am working off of a translation here.)  Does anyone have any idea what proof Hilbert could be referring to?
 A: It could be Kronecker's determination of the sign of the Gauss sum by means of Cauchy's theorem. Already Gauss noted that the determination of the sign implies the law of quadratic reciprocity.
In response to the request for references:
Leopold Kronecker: Summirung der Gauss'schen Reihen ... J. Reine Angew. Math. 105 (1889), 267-268.
Also in volume 4 of his Werke, 297-300. (This was where I xeroxed it, so I can vouch for the page numbers, I have the pages in front of me right now).
Also in Landau's Elementare Zahlentheorie (together with two others, by Mertens and Schur),
near the end of the book.
Also supposed to be in Ayoub: Introduction to the Analytic Theory of Numbers, but I am not familiar with his book, so I cannot vouch for this.
There is a later determination of the sign of the Gauss sum by contour integration, due to Mordell, which is quite accessible; it is in Chandrasekharan's Introduction to Analytic Number Theory, page 35--39. Chandrasekharan does a more general case.
Now, I have not claimed that Kronecker's proof was the one that Hilbert was thinking of. I cannot read the mind of a dead man (nor that of a living one).
A: A will add a few comments on analytic proofs of quadratic reciprocity. The first one is due to Dirichlet  in 1835, using the Poisson summation formula but not Cauchy's theorem nor the functional equation of the theta series. The functional equation for the theta series used in Cauchy's 1840 analytic proof was first established by Jacobi. He used neither Poisson summation nor Cauchy's theorem, but derived the functional equation by formula manipulation within the framework of his theory of elliptic functions. The functional equation of the theta series can be established without Cauchy's theorem, by the Poisson summation formula, or by the Euler-Maclaurin summation formula and Fourier analysis. It can also be established by the Plana summation formula, and there is a direct proof too of quadratic reciprocity by the Plana summation formula.
The early work on elliptic functions by Abel and Jacobi made no use of the concept of analytic function or Cauchy's theorem.
A: $\def\FF{\mathbb{F}}$I'm just guessing, but I would have thought it was the following: Hilbert reciprocity for function fields can be deduced from Weil reciprocity. Weil reciprocity is the following statement: Let $X$ be a complete curve over an algebraically closed field $k$. For any point $x \in X$ and nonzero meromorphic functions $f$ and $g$, define $(f,g)_x = (-1)^{(\mathrm{ord}_x f)(\mathrm{ord}_x g)}(f^{\mathrm{ord}_x g}/g^{\mathrm{ord}_x f})(x)$. Then $\prod_{x \in X} (f,g)_x=1$.
See here and here for the connection.
Now, over $\mathbb{C}$, we can prove Weil reciprocity as follows: Choose a path $\delta$ connecting $0$ to $\infty$ in $\mathbb{CP}^1$ and avoiding the critical values of $f$. For simplicity, let us assume $f$ has simple zeroes and poles $\zeta^{\pm}_1$, $\zeta^{\pm}_2$, ..., $\zeta^{\pm}_n$. Set $\gamma = f^{-1}(\delta)$. Then $\gamma$ is the union of $\deg(f)$ closed line segments. After reordering, we may assume  $\zeta^+_i$ is joined to $\zeta^-_i$, say by $\gamma_i$. 
We can define $\log(f)$ on $X \setminus \gamma$, by composing $f$ with a branch of $\log$ on $\mathbb{CP}^1 \setminus \delta$. The differential form $\omega:= \tfrac{1}{2 \pi i} \log(f) \tfrac{dg}{g}$ therefore makes sense on $X \setminus (\gamma \cup g^{-1}(\{ 0,\infty \}))$. If we integrate $\omega$ on little contours around the zeroes and poles of $g$, we get $\sum_{x \in X} \mathrm{ord}_x(g) \log(f(x))$. 
On the other hand, if we integrate around a tubular neighborhood of $\gamma_i$, we pick up $\int_{\gamma_i} \tfrac{dg}{g} = \log(g(\zeta^{+}_i) - \log(g(\zeta^-_i))$ for some branch of $\log$. Summing on $i$, this is
$\sum_{x \in X} \mathrm{ord}_x(f) \log(g(x))$
The sum of the contours around the zeroes of $f$ is homologous to the sum over the neighborhoods of the $\gamma_i$, so we deduce
$$\sum_{x \in X} \mathrm{ord}_x(g) \log(f(x)) = \sum_{x \in X} \mathrm{ord}_x(f) \log(g(x))$$
and exponentiating gives the result.
