Trigonometric sums related to the Verlinde formula

Original question (see also the revised, possibly simpler, version below): Let g>1,r>1 be integers. Playing around with the Verlinde formula (see below), I came across the expression
r1n=1sin(πn/r)22g(e2πin2/r1).

My goal is to reduce the complexity in r of this expression; that is, to find a closed form of the sum avoiding the dependence on r in the number of summands. Is this possible? Here’s a related example:

The Verlinde formula, which e.g. has applications in conformal field theory, algebraic geometry, and quantum topology, is
(r/2)g1r1n=1sin(πn/r)22g.
In this case, one can use a trick by Szenes to reduce the complexity of the sum: The sum can be written as
r1n=1f(zn).
where zn=eπin/r, for a suitable meromorphic function f:CC having poles only at 1 and 1. Now the trick essentially is to find a suitable meromorphic form μr having poles at 2r‘th roots of unity and apply the Residue Theorem to fμk to rewrite the above sum as
r1n=1f(zn)=Resz=1fμr,
which then turns out to be a polynomial in r of degree 2g2.

This trick doesn’t seem to apply to my slightly more complicated sum though. Another possibility might be to somehow rewrite the sum as a Gauss sum, but that doesn’t quite seem to work either.

“Revised” question: So, maybe the question above does not have a straightforward answer, but I believe it might suffice to be able to work out the following (at least, it’s a similar problem). Say we just have a sum like
r1n=1eπin2/(2r)
as above (almost, anyway). Then we may apply a quadratic reciprocity theorem to simplify matters. But say now that we throw in a power of n to get something like
r1n=1nkeπin2/(2r),
for k>0. Can sums like these be treated in a similar manner as the quadratic Gauss sum above (perhaps just in special cases like k=1, or k=2); can we somehow describe the large r asymptotics? Standard tricks in this field seem to involve summation by parts and the Euler–Maclaurin formula but it doesn’t seem to quite work out. For example, in the case k=1, summation by parts (or elementary combinatorial considerations) will imply
r1n=1neπin2/(2r)=(r1)r1n=1eπin2/(2r)r1j=1jn=1eπin2/(2r).
Now, the first term is simple to handle as mentioned above, but the second one seems to be worse. Any suggestions?

Answer

Proposition 1 answers the “revised” question and Proposition 2 the original one. For completeness we give a self-contained proof of Proposition 2.

Proposition 1

Let nN and let p(x)=αx2+βx+γ be a polynomial with real coefficients where α>0 such that p(n)>0 and p(n+1)<1. Then for kZ if n>0 and for kN if n=0 we have
12(nke2πip(n)+(n+1)ke2πip(n+1))=
e2πip(n+1)ϕk(n+1,p
where \phi_k is the function defined on A\times ]0,\infty[\times]0,1[ with A=\mathbb{N} or \mathbb{N}^* according to k\geq 0 or k<0 by

\phi_k(m,\lambda,\mu)=\frac{i}{2} \int_0^{\infty}\left((m-ix)^k e^{2\pi( \mu-\frac{1}{2})x}-(m+ix)^k e^{-2\pi( \mu-\frac{1}{2})x}\right)\text{csch}(\pi x) e^{-i\pi \lambda x^2}\,dx\,.

The main ingredient in the proof of Proposition 1 is Lemma 1 which is proved in my paper "Sommes exponentielles, splines quadratiques et fonction zêta de Riemann" published in the "Comptes-rendus de l'Académie des sciences" in 2001 and avalaible at this address

http://math.heig-vd.ch/fr-ch/Recherche/Recherches/Philippe_Blanc_novembre_2000.pdf

A detailed version is avalaible at this address

http://math.heig-vd.ch/fr-ch/Recherche/Recherches/Philippe_Blanc_mars_2001.pdf

Lemma 1

Let n be an integer, p(x)=\alpha x^2+\beta x +\gamma be a polynomial with real coefficients where \alpha >0 and let z(\cdot) be the unique function satisfying p'(z(y))=y for all y \in \mathbb{R}. Then

\frac{1}{2}(e^{2\pi i p(n)}+e^{2\pi i p(n+1)})=\frac{e^{i\frac{\pi}{4}}}{\sqrt{p''(n+\frac{1}{2})}}\sum_{\lfloor p'(n)\rfloor +1}^{\lfloor p'(n+1)\rfloor }e^{2\pi i(p(z(k))-kz(k))}+


e^{2\pi i p(n+1)}\phi (p''(n+\frac{1}{2}),\{p'(n+1)\})-e^{2\pi i p(n)}\phi (p''(n+\frac{1}{2}),\{p'(n)\})

where \lfloor \cdot \rfloor and \{\cdot\} denote respectively the floor and fractional part functions and \phi is the function defined on ]0,\infty[\times[0,1] by

\phi(\lambda,\mu)=i \int_0^{\infty}\sinh (2\pi( \mu-\frac{1}{2})x)\text{csch}(\pi x) e^{-i\pi \lambda x^2}\,dx\,.\hspace{3mm}\Box

Proof of Proposition 1

With the assumptions on the derivatives of p the sum which appears in Lemma 1 is void and we have

\frac{1}{2}(e^{2\pi i p(n)}+e^{2\pi i p(n+1)})=


e^{2\pi i p(n+1)}\phi(p''(n+\frac{1}{2}),p'(n+1))-e^{2\pi i p(n)}\phi(p''(n+\frac{1}{2}),p'(n)) \tag{1}

which proves the case k=0.

Considering the terms of (1) as a function of \beta and differentiating k times with respect to \beta we get the proposition for k>0.

Then we replace the polynomial p by p_z(x)=p(x)+zx where z\in \mathbb{C}. The left hand side of (1), considered as a function of z, is holomorphic. The right hand side of (1) is also holomorphic in the band B=\{z\in \mathbb{C}\vert -p'(n)< \Re z<1-p'(n+1) \}. Since identity (1) holds for real z\in B, it holds in B. We set p_{it_1}(x)=p(x)+it_1 x in identity (1), integrate a first time with respect to t_1 on the interval [t_2,\infty[, a second time with respect to t_2 on the interval [t_3,\infty[,..., and finally integrate a k-th time with respect to t_k on the interval [0,\infty[ to complete the proof in the case k<0\,.\hspace{3mm}\Box

The functions \phi_k extend by continuity on A\times ]0,\infty[ \times [0,1]. Choosing, for example, k=1, p(x)=\frac{x^2}{4r} in the identity of Proposition 1 and summing these identities from n=0 to r-1 we get

\sum_{n=1}^{r-1} n e^{\pi i \frac{n^2}{2r}}=-\frac{1}{2}re^{\pi i \frac{r}{2}}+e^{\pi i \frac{r}{2}}\phi_1(r,\frac{1}{2r},\frac{1}{2})-\phi_1(0,\frac{1}{2r},0)=


=\left ( -\frac{1}{2}e^{\pi i \frac{r}{2}}+\frac{i}{\pi}\right )r-\int_0^{\infty}\frac{xe^{-\pi x}}{\sinh \pi x}e^{-i \pi \frac{1}{2r}x^2 }\,dx+
\left ( \int_0^{\infty}\frac{x}{\sinh \pi x}e^{-i\pi \frac{1}{2r} x^2}\,dx\right )e^{\pi i \frac{r}{2}}

which implies that

\sum_{n=1}^{r-1} n e^{\pi i \frac{n^2}{2r}}=\left (-\frac{1}{2}e^{\pi i \frac{r}{2}}+\frac{i}{\pi}\right )r-\frac{1}{12}+\frac{1}{4}e^{\pi i \frac{r}{2}}+O(\frac{1}{r})\,.

Now for x\in]0,\pi[ and an integer g>1 we have

(\sin x)^{2-2g}=\sum_{k=1-g}^{\infty}\alpha_{2k}x^{2k}

and Proposition 1 suggests the following result.

Proposition 2

Let r,\,g and n be integers such that r>1, g>1 and 1\leq n \leq \frac{r}{2}-1. Then

\frac{1}{2}\left ( (\sin \pi \frac{n}{r})^{2-2g} e^{2\pi i \frac{n^2}{r}}+(\sin \pi \frac{n+1}{r})^{2-2g} e^{2\pi i \frac{(n+1)^2}{r}}\right )=
e^{2\pi i \frac{(n+1)^2}{r}}\Psi_{r,g}(n+1)-e^{2\pi i \frac{n^2}{r}}\Psi_{r,g}(n)

where \Psi_{r,g} is the function defined on \{1,2,\ldots,\lfloor \frac{r}{2}\rfloor\} by

\Psi_{r,g}(m)=


\frac{i}{2} \int_0^{\infty}\left ((\sin\pi \frac{m-ix}{r})^{2-2g}
e^{2\pi( \frac{2m}{r}-\frac{1}{2})x}-(\sin\pi \frac{m+ix}{r})^{2-2g} e^{-2\pi( \frac{2m}{r}-\frac{1}{2})x}\right )
\,\text{csch}(\pi x) e^{-i \pi \frac{2}{r} x^2}
\, dx

Proof of Proposition 2

Introducing the function f(x)=(\sin x)^{2-2g} we have

\int_0^{\infty}\left ( f(\pi \frac{m-ix}{r})
e^{2\pi( \frac{2m}{r}-\frac{1}{2})x}-f(\pi \frac{m+ix}{r}) e^{-2\pi( \frac{2m}{r}-\frac{1}{2})x}\right ) \,\text{csch}(\pi x) e^{-i \pi \frac{2}{r} x^2}
\, dx =


\text {PV}\int_{-\infty}^{\infty}f(\pi \frac{m-ix}{r})
e^{2\pi( \frac{2m}{r}-\frac{1}{2})x} \,\text{csch}(\pi x) e^{-i \pi \frac{2}{r} x^2}
\, dx \tag{2}

for m\in \{1,2,\ldots,\lfloor \frac{r}{2}\rfloor\}.
Now we compute

\text {PV}\int\limits_{C_R}f(\pi \frac{n-iz}{r})
e^{2\pi( \frac{2n}{r}-\frac{1}{2})z} \,\text{csch}(\pi z) e^{-i \pi \frac{2}{r} z^2}\,dz

where C_R is the boundary of the rectangle with vertices -R, R, R+i and -R+i and taking the limit as R \to \infty and using the residue theorem we get

\text {PV}\int_{-\infty}^{\infty}f(\pi \frac{n-ix}{r})
e^{2\pi( \frac{2n}{r}-\frac{1}{2})x} \,\text{csch}(\pi x) e^{-i \pi \frac{2}{r} x^2}
\, dx +


\text {PV}\int_{\infty}^{-\infty}f(\pi \frac{n-i(i+x)}{r})
e^{2\pi( \frac{2n}{r}-\frac{1}{2})(i+x)} \,\text{csch}(\pi (i+x)) e^{-i \pi \frac{2}{r} (i+x)^2}
\, dx =


\pi i \left( \frac{f(\pi \frac{n}{r})}{\pi}+e^{2\pi i \frac{2n+1}{r}}\frac{f(\pi \frac{n+1}{r})}{\pi}\right )\,.

Finally we multiply this identity by -\frac{i}{2}e^{2\pi i \frac{n^2}{r}} and we complete the proof observing that

\text {PV}\int_{\infty}^{-\infty}f(\pi \frac{n-i(i+x)}{r})
e^{2\pi( \frac{2n}{r}-\frac{1}{2})(i+x)} \,\text{csch}(\pi (i+x)) e^{-i \pi \frac{2}{r} (i+x)^2}
\, dx =


-e^{2\pi i \frac{2n+1}{r}}\text {PV}\int_{-\infty}^{\infty}f(\pi \frac{n+1-ix)}{r})
e^{2\pi( \frac{2(n+1)}{r}-\frac{1}{2})x} \,\text{csch}(\pi x) e^{-i \pi \frac{2}{r} x^2}
\, dx

and using relation (2).\hspace{3mm}\Box

Coming back to the original question in the case g=2, assuming r odd for simplicity and summing the identities of Proposition 2 from n=1 to \frac{r-3} {2} we have

\sum_{n=1}^{r-1} \left (\sin \pi \frac{n}{r}\right )^{-2} \left (e^{2\pi i \frac{n^2}{r}}-1\right )=2\sum_{n=1}^{\frac{r-1}{2}} \left ( \sin \pi \frac{n}{r}\right )^{-2} \left (e^{2\pi i \frac{n^2}{r}}-1\right )=


2\left ( \frac{1}{2}\left (\sin \pi \frac{1}{r}\right )^{-2} e^{2\pi i \frac{1}{r}}+\frac{1}{2}\left ( \sin \pi \frac{r-1}{2r}\right )^{-2} e^{2\pi i \frac{(r-1)^2}{4r}}\right ) +


2\left ( e^{2\pi i \frac{(r-1)^2}{4r}}\Psi_{r,2}(\frac{r-1}{2})-e^{2\pi i \frac{1}{r}}\Psi_{r,2}(1)-\sum_{n=1}^{\frac{r-1}{2}} \left ( \sin \pi \frac{n}{r}\right ) ^{-2}\right).

Proposition 3


\Psi_{r,2}(1)=\frac{1}{\pi^2}\left ( \frac{1}{2}-\frac{\pi^2}{6}\right ) r^2+\frac{1}{\pi}(1-i)r^{\frac{3}{2}}+O(r)\,.

Proof of Proposition 3

By introducing the real-valued functions g_{re} and g_{im} defined by the relation

\left ( \sin \pi \frac{1 + ix}{r}\right )^{-2}=\frac{r^2}{\pi ^2}\frac{1}{(1+ i x)^2} +g_{re}(x)+ i g_{im}(x)

and setting \displaystyle \mu=\frac{2}{r} for ease of notations we have

\Psi_{r,2}(1)=\frac{r^2}{\pi ^2}\Theta +\Phi

where

\Theta =\frac{i}{2} \int_0^{\infty}\left (\frac{1}{(1-ix)^2} e^{2\pi( \mu-\frac{1}{2})x}-\frac{1}{(1+ix)^2} e^{-2\pi( \mu-\frac{1}{2})x}\right )\,\text{csch}(\pi x) e^{-i\pi \mu x^2}\,dx \,=A+B

with

A=i \int_0^{\infty}\frac{1-x^2}{(1+x^2)^2} \sinh(2\pi( \mu-\frac{1}{2})x)\,\text{csch}(\pi x) e^{-i\pi \mu x^2}\,dx \,,


B=-\int_0^{\infty}\frac{2x}{(1+x^2)^2} \cosh(2\pi( \mu-\frac{1}{2})x)\,\text{csch}(\pi x) e^{-i\pi \mu x^2}\,dx

and

\Phi=i \int_0^{\infty}\left ( g_{re}(x)\sinh ( 2\pi( \mu-\frac{1}{2})x)-i g_{im}(x)\cosh (2\pi( \mu-\frac{1}{2})x)\right )\,\text{csch}(\pi x) e^{-i\pi \mu x^2}\,dx\,.

As \displaystyle \int_0^t e^{-i\pi \mu x^2}\,dx=O(\mu^{-\frac{1}{2}}) we can use the second mean value theorem to check that \displaystyle \Phi=O(\mu^{-\frac{1}{2}}).
Further we use the identity

\sinh(2\pi (\mu-\frac{1}{2})x)\text{csch} \pi x =
-e^{-2\pi \mu x}+e^{-\pi x}\sinh(2\pi \mu x)\text{csch} \pi x

to write A=A_1 + A_2 and we bound the modulus of A_2 by the integral of the modulus to get A_2=O(\mu ).
Using an integration by parts we have

A_1 = -i\int_0^{\infty}\frac{1-x^2}{(1+x^2)^2}e^{-2\pi \mu x} e^{-i\pi \mu x^2}\,dx=


=-2\pi \mu i \int_0^{\infty}\frac{x}{1+x^2}e^{-2\pi \mu x}e^{-i\pi \mu x^2}\,dx+2\pi \mu \int_0^{\infty}\frac{x^2}{1+x^2}e^{-2\pi \mu x}e^{-i\pi \mu x^2}\,dx=


2\pi \mu \int_0^{\infty}e^{-2\pi \mu x}e^{-i\pi \mu x^2}\,dx-2\pi \mu i \int_0^{\infty}\frac{x}{1+x^2}e^{-2\pi \mu x}e^{-i\pi \mu x^2}\,dx-


2\pi \mu \int_0^{\infty} \frac{1}{1+x^2} e^{-2\pi \mu x}e^{-i\pi \mu x^2}\,dx\,.

We make use of relations (8.256.3) and (8.256.4) of Gradshteyn and Ryzhik to conclude that

A_1=\pi e^{-i\frac{\pi}{4}} \mu^{\frac{1}{2}}-2\pi \mu i \int_0^{\infty}\frac{x}{1+x^2}e^{-2\pi \mu x}e^{-i\pi \mu x^2}\,dx+O(\mu)\,.

Similarly we use the identity

\cosh(2\pi (\mu-\frac{1}{2})x)\text{csch} \pi x\ =\coth(\pi x)+2\sinh(\pi (\mu-1) x)\sinh(\pi \mu x)\text{csch} \pi x

to write B=B_1+B_2 where B_2=O(\mu).
We have, using the identity \displaystyle \coth \pi x =1 +\frac{2}{e^{2\pi x}-1}, relation (3.415.2) of Gradshteyn and Ryzhik and an integration by parts
B_1=-\int_0^{\infty}\frac{2x}{(1+x^2)^2}\coth (\pi x) e^{-i\pi \mu x^2}\,dx=

-\int_0^{\infty}\frac{2x}{(1+x^2)^2}\coth (\pi x)\,dx -\int_0^{\infty}\frac{2x}{(1+x^2)^2}(e^{-i\pi \mu x^2}-1)\,dx +O(\mu)=


\frac{1}{2}-\frac{\pi^2}{6}+2\pi \mu i \int_0^{\infty}\frac{x}{1+x^2}e^{-i\pi \mu x^2}\,dx+O(\mu)\,.

Finally

\Theta=A_1+B_1+O(\mu)=\frac{1}{2}-\frac{\pi^2}{6}+\pi e^{-i\frac{\pi}{4}} \mu^{\frac{1}{2}}+2\pi \mu i \int_0^{\infty}\frac{x}{1+x^2}(1-e^{-2\pi \mu x})e^{-i\pi \mu x^2}\,dx+O(\mu)

and using the second mean value theorem we get

\Psi_{r,2}(1)=\frac{r^2}{\pi ^2}\left ( \frac{1}{2}-\frac{\pi^2}{6}+\pi (1-i) r^{-\frac{1}{2}}+O(\frac{1}{r})\right )=


\frac{1}{\pi^2}\left ( \frac{1}{2}-\frac{\pi^2}{6}\right ) r^2+\frac{1}{\pi}(1-i)r^{\frac{3}{2}}+O(r)\,.\hspace{3mm}\Box

Proposition 3 together with the fact that
\displaystyle \Psi_{r,2}(\frac{r-1}{2})=O(r) imply
\sum_{n=1}^{r-1} \left ( \sin \pi \frac{n}{r}\right ) ^{-2} \left ( e^{2\pi i \frac{n^2}{r}}-1\right ) =\frac{2}{\pi}(-1+i)r^{\frac{3}{2}}+O(r)\,.

Note that it is easy to prove that the previous sum is a \displaystyle O(r^{\frac{3}{2}}) by using the bound \displaystyle \vert e^{it}-1 \vert \leq \min (t,2).

Attribution
Source : Link , Question Author : fuglede , Answer Author : Philippe Blanc

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