Question: Let $s(p,q)=\sum_{i=1}^{q-1}((i/q))((pi/q))$ where (p,q)=1 and $((x))=x-[x]-1/2$ for $x\notin Z$. I want to prove that $s(p,q)+s(q,p)=(p/q+\frac{1}{pq}+q/p)/12-1/4$ using at least one of the following four methods.

i. I tried proving Dedekind Reciprocity using Riemann-Stieltjes integration. In his 1982 journal article, ‘Sums involving the greatest integer function and Riemann-Stieltjes integration’ by Bruce C. Berndt, he recommends using Riemann-Stieltjes integration to prove identities involving sums of the greatest integer function where it would be difficult to do so using lattice-point diagrams. You can read it online at http://gdz.sub.uni-goettingen.de/dms/load/img/?PPN=GDZPPN002200031. He proves a result by Carlitz using an identity needed to prove Dedekind Reciprocity, but does not prove that identity. I am open to any ideas, but it seemed like Riemann-Stieltjes integration can only be applied in a select few cases.

ii. I tried proving Dedekind Reciprocity using lattice-point diagrams, but it got very messy. Since Berndt did not prove it using Riemann-Stieltjes integration, I doubt it can be done using lattice-points. I welcome any ideas for using lattice-point diagrams in a critical part (if not the main part) of the proof.

iii. I tried proving Dedekind Reciprocity using the Fourier series for the sawtooth-wave function. I actually made some progress with this approach. Since $-((x))=\sum_{n=1}^{\infty}sin(2n{\pi}x)/(n\pi)$ for ALL REAL x, we consider the function $((i/q))((pi/q))$ where (p,q)=1. Since ((i/q)) simplifies to i/q-1/2 for all i in the sum, I can evaluate all sums in the product using elementary methods except for $\sum_{i=1}^{q-1}(i/q)((pi/q))+\sum_{i=1}^{p-1}(i/p)((qi/p))$. Since $i((pi/q))/q=-\sum_{n=1}^{\infty}isin(2n{\pi}i/q)/(qn\pi)$, we start with the finite sum $\sum_{k=1}^{q-1}kx^k=\frac{qx^q}{x-1}-\frac{x(x^q-1)}{(x-1)^2}$. Since $ksin(2n{\pi}kp/q)/(qn\pi)=k((e^{i2n{\pi}p/q})^k-(e^{-i2n{\pi}p/q})^k)/(2qn{\pi}i)$ and $\sum_{n=1}^{\infty}\sum_{k=1}^{q-1}k((e^{i2n{\pi}p/q})^k-(e^{-i2n{\pi}p/q})^k)/(2qn{\pi}i)=\sum_{n=1}^{\infty}(1/2-1/(1-e^{i2n{\pi}p/q}))/(n{\pi}i)$, $\sum_{i=1}^{q-1}(i/q)((pi/q))+\sum_{i=1}^{p-1}(i/p)((qi/p))=-\sum_{n=1}^{\infty}(1-1/(1-e^{i2n{\pi}p/q})-1/(1-e^{i2n{\pi}q/p}))/(n{\pi}i)$. When converting the sum back into trigonometric functions, the sum simplifies to $\sum_{n=1}^{\infty}(cot(n{\pi}p/q)+cot(n{\pi}q/p))/(2n\pi)=\sum_{n=1}^{\infty}\frac{sin(n{\pi}(p/q+q/p))}{2n{\pi}sin(n{\pi}p/q)sin(n{\pi}q/p)}$. How can I evaluate these sums? Can these sums be easily evaluated using Poisson summation or some other analytic method?

iv. I found a decent proof of Dedekind Reciprocity in a 1953 journal article by Carlitz; however, the disadvantage of using it is that I would have to introduce Lagrange polynomials. You can read it online at http://projecteuclid.org/DPubS/Repository/1.0/Disseminate?view=body&id=pdf_1&handle=euclid.pjm/1103051326. The first step in the proof requires a proof that $-((r/k))=\frac{1}{2k}+\frac{1}{k}\sum_{s=1}^{k-1}\frac{p^{-rs}}{(1-p^s)}$ where $p=e^{2{\pi}i/k}$ and $r,k\in Z$. The article recommended Mobius inversion, but this relation looks simple enough that I want to prove it by brute force. I am currently working on it and welcome any ideas.

Background: I am writing a subsection on Dedekind Reciprocity for my online book, *A Mathematical Analysis of the Greatest Integer Function*. The content of my book can be more accurately summarized as a discussion of a variety of integer-intensive topics in number theory and analysis. My objective is to develop a lot of theory on integer functions in the beginning and then use the theory to discuss these integer-intensive topics and prove important results associated with them using a powerful, direct, and unified approach. In summary, I am attempting to devise a proof of Dedekind Reciprocity that uses integer functions in an advanced way and is otherwise elementary. However, if such a proof already exists, then I would be content with inserting it in my work and citing the source.