I would like to estimate the absolute value of the following difference

$$ \Delta(L) = \sum_{\alpha=-L+1}^L \frac{1}{1+2 L} e^{i t \sec^2\left(\pi\frac{\alpha - 1/2}{2 L+1}\right)} - \int_{-\frac{1}{2}}^\frac{1}{2} e^{i t \sec^2(\pi x)}dx $$

where $L$ is an integer, for $L \to \infty$. I am assuming $t$ has a positive imaginary part.

Numerically, I have evidence that the error decays like $|\Delta(L)| \approx c L^{-d} e^{- a L^b}$, with $b$ approximately $\frac{1}{2}$ or $\frac{2}{3}$. I would like to find $b$ and $a$ (I'm not so interested in $d$ and $c$).

I tried to use the Euler-Maclaurin formula to estimate the error, but all derivatives of the boundary function $e^{i t \sec^2\left(\pi\frac{x - 1/2}{2 L+1}\right)}$ vanish in the strict $L \to \infty$ limit, and I'm not sure how to resum them for large $L$.

  • $\begingroup$ Since you have the means to investigate the error numerically, maybe you could try linear regression on $$\log(- \log | \Delta|)$$ Then you can estimate $a$ and $b$ if the dependence indeed has the form you suggest $\endgroup$ – Yuriy S Nov 16 '18 at 21:05
  • $\begingroup$ @YuriyS that's a good point, I should try that more, however since the coefficient $a$ depends on $t$ that's not too easy. By the way why you study the double log? $\endgroup$ – chubecca Nov 16 '18 at 21:06
  • $\begingroup$ @\chubecca, from your claim we have $-\log |\Delta|=a L^b$. Then to find both $a$ and $b$ we can plot $\log(a L^b)=\log a + b \log L$ vs $\log L$, which should look like a straight line. With least squares you can find the parameters $\endgroup$ – Yuriy S Nov 16 '18 at 21:11
  • $\begingroup$ @YuriyS sorry, I edited the post to make my statement more accurate $\endgroup$ – chubecca Nov 16 '18 at 21:14
  • $\begingroup$ The exponential behavior will still be more prominent for large $L$, so you could try my suggestion. $\endgroup$ – Yuriy S Nov 16 '18 at 21:22

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