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

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
In this case, one can use a trick by Szenes to reduce the complexity of the sum: The sum can be written as
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
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
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
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
Now, the first term is simple to handle as mentioned above, but the second one seems to be worse. Any suggestions?


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
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

A detailed version is avalaible at this address

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


\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


\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


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


\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)\,.


\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).

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

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