Using the Euler-Maclaurin summation formula, we can prove that$$\log η(τ)=\frac{iπτ}{12}-\frac{iπ}{12τ}-\log\sqrt{-iτ}-\int_0^∞p(y)\left[\frac{2πiτ}{e^{-2πiτy}-1}+\frac1y\right]\,\mathrm dy,$$where $\eta(\tau)$ is the Dedekind eta function, and $p(y)=\left\{y\right\}-\frac{1}{2}$ ($\{y\}$ being the fractional part of $y$). My question: How do we use this formula to deduce the transformation formula: $$\eta\left(\frac{a\tau+b}{c\tau+d}\right)=\epsilon(a,b,c,d)[-i(c\tau+d)]^{\frac{1}{2}}\eta(\tau),$$ where the integers $a,b,c$ and $d$ satisfy $c>0$ and $ad-bc=1$, and $\epsilon(a,b,c,d)$ is given by: $$\epsilon(a,b,c,d)=\exp \left[i\pi\left(\frac{a+b}{12c}-s(d,c)-\frac{1}{4}\right)\right],$$ where $s(d,c)$ is the Dedekind sum, and is given by: $$s(d,c)=\sum_{n=1}^{c-1}\frac{n}{c}\left(\frac{dn}{c}-\left \lfloor \frac{dn}{c} \right \rfloor -\frac{1}{2}\right).$$ My gut feeling tells me that the Mittag-Leffler expansion: $$\frac{1}{e^{z}-1}=\frac{1}{z}-\frac{1}{2}+2z\sum_{n=1}^{\infty}\frac{1}{z^{2}+4\pi^{2}n^{2}}$$ is key here, but I have no idea how to proceed!
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1Why do you think that it can be done? Where have you looked? Did you look in Apostol, Modular Functions and Dirichlet Series in Number Theory, Chapter 3 "The Dedekind eta function, its functional equation, and Dedekind sums"? – Somos Apr 10 '19 at 12:04
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I am aware of the available proofs of transformation formula of $\eta(\tau)$, but i haven't seen this formula anywhere ... i have derived it myself. Using the Fourier expansion of fractional part function, i was able to prove $\eta\left(\frac{-1}{\tau}\right)=\sqrt{-i\tau}\eta(\eta)$. But i am curious how this formula could be used to prove the 'full' transformation formula of $\eta(\tau)$. – Mohammad Al Jamal Apr 10 '19 at 12:16
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The modular group is generated by $\tau \mapsto \tau+1, \tau \mapsto -1/\tau$, once you know how $\eta$ transforms for those you know how it transforms for the full modular group. – reuns Apr 13 '19 at 02:56