It looks like a relatively straight-forward application of the residue theorem. You need to find all the poles inside your rectangle, and then evaluate the residues of f(z) at each of those poles. Can you go from here?
I'm trying to prove the reciprocity law for quadratic Gauss sums when m=1(mod 4) and need help showing something. Could anyone help? It's been a while since I done any residue calculus.
Let C be the rectangular contour with vertices, iR, -iR, 1/2m+iR and 1/2m-iR. It also has a semicircular indentation of radius epsilon (< 1) at 0.
Clearly f is analytic everywhere, except for a first order pole at each integer, but I am stumped at how to get the required result.
I think its the fact that m=1(mod 4) that is throwing me.
When m=1, there are no poles in the rectangle and so the integral is zero. For m=5, there is a pole at 1 and 2. At m=9 there are poles at 1, 2, 3 and 4 and so on.
Do I need to continue in this way for infinitely many poles, or have I got the wrong idea here?
The poles of occur precisely when The numerator is never zero, so we don't have any zero-over-zero limits to worry about. Using Euler's formula, we have
For this equation to hold, and This occurs, as you have said, precisely when
Now, if I examine the contour you've chosen, there is one question I would ask: the indent at the origin: is that to the right or to the left? If it's to the right, then I would argue that the integral contrary to the hypothesis. All the poles are integers. If I look at the real axis and scan the interval for integers , I don't think there are any in that region, and therefore no poles in that region. Therefore, the contour integral of the analytic function around a simple closed contour is zero, by the Cauchy-Goursat theorem.
Are you sure you've stated everything in the problem accurately?
Everything you have mentioned is correct. The indent is to the right and so excludes 0.
I'm working from Apostols "Analytic Number Theory", where he discusses the transformation f(z+1). If I use this, then the zero pole would be inside the contour, but would still not give the result.
Therefore by Cauchy's theorem the integral round the contour is equal to .
To get that result into the form required, notice that . (The reason for that is that , and e to the power times an integer is equal to 1.) Thus the term in the above sum can be split into two, as , and then you can rearrange the sum so that it goes from 1 to m–1 instead of from 1 to (1/2)(m–1).