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Are there asymptotics, or even a closed form, for the following series

$$ \sum_{k = 0}^\infty e^{2 \pi i \sqrt{k^2 + (d-1) k} } \left( \binom{d+k}{k} - \binom{d+k-2}{k-2} \right) G_{\frac{d-1}{2},k}(t) $$

where $G_{\frac{d-1}{2},k}$ is the Gegenbauer polynomial of degree $k$, orthogonal with respect to the weight function $(1 - t^2)^{d/2-1}$ on $[-1,1]$ and with $G_{\frac{d-1}{2},k}(1) = 1$ for all $k$.

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  • $\begingroup$ just to be sure: in your notation, is $G_{1,2}(x)$ equal to $4x$ or to $4x^2-1$? $\endgroup$
    – Carlo Beenakker
    Commented Jun 25 at 20:47
  • $\begingroup$ I've added clarification (and corrected a couple typos). The normalization chosen is so that $G_{1,2}$ is a quadratic polynomial with $G_{1,2}(1) = 1$. So we have $G_{1,2}(t) = 2t^2 - 1$. $\endgroup$
    – Jacob Denson
    Commented Jun 25 at 23:42
  • $\begingroup$ my confusion remains; the unnormalised Gegenbauer polynomial $C_2^{(1)}(t)=4t^2-1$, so when you normalise you do not get $2t^2-1$ ... is this a typo? $\endgroup$
    – Carlo Beenakker
    Commented Jun 26 at 10:56
  • $\begingroup$ Sorry yes there was a mistake here. $2t^2 - 1$ is $G_{0,2}$, whereas $(1/3)(4t^2 - 1)$ is $G_{1,2}$ $\endgroup$
    – Jacob Denson
    Commented Jun 26 at 14:58

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A simplification of the expression that may be helpful: $$e^{2 \pi i \sqrt{k^2 + (d-1) k} } \left( \binom{d+k}{k} - \binom{d+k-2}{k-2} \right) G_{\frac{d-1}{2},k}(t)=$$ $$=e^{2 \pi i \sqrt{k^2 + (d-1) k} }\frac{ d+2 k-1 }{d-1}C_k^{\left(\frac{d-1}{2}\right)}(t),$$ with $C_k^{(\alpha)}$ the regular Gegenbauer polynomial, as defined on Wikipedia without the normalisation of the OP. The two are related by $$G_{\frac{d-1}{2},k}(t)=\frac{k!}{(d-1)_k}\,C_k^{\left(\frac{d-1}{2}\right)}(t).$$ So the OP seeks the sum $$F_d(t)=\frac{1}{d-1}\sum_{k=0}^\infty e^{2 \pi i \sqrt{k^2 + (d-1) k} }(d+2 k-1 )C_k^{\left(\frac{d-1}{2}\right)}(t).$$ A closed-form answer seems unlikely.
For $d=2$ one can work with Legendre polynomials, $$F_2(t)=\sum_{k=0}^\infty e^{2\pi i \sqrt{k (k+1)}} (2 k+1) P_k(t),$$ but also that expression is not conducive to a closed-form evaluation.

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    $\begingroup$ The expression comes from a singular integral, so I would expect the function to have a singularity at $t=0$, likely of order $d$. $\endgroup$
    – Jacob Denson
    Commented Jun 26 at 13:46
  • $\begingroup$ indeed, the $t=0$ limit is singular; perhaps showing the integral expression will help for the asymptotics? $\endgroup$
    – Carlo Beenakker
    Commented Jun 26 at 19:35
  • $\begingroup$ There isn't exactly an explicit expression for the singular integral, I just know this quantity must define a pseudodifferential operator of order zero (and thus be the convolution kernel of a singular integral). The quantity arises from studying the operator $e^{2 \pi i \sqrt{-\Delta}}$ on the sphere (a paper from 1978 due to Guillemin calls this the return operator). $\endgroup$
    – Jacob Denson
    Commented Jun 29 at 23:08

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