# Convergence of sum to integral

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 $$L$$.

• 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 – Yuriy S Nov 16 '18 at 21:05
• @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? – chubecca Nov 16 '18 at 21:06
• @\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 – Yuriy S Nov 16 '18 at 21:11
• @YuriyS sorry, I edited the post to make my statement more accurate – chubecca Nov 16 '18 at 21:14
• The exponential behavior will still be more prominent for large $L$, so you could try my suggestion. – Yuriy S Nov 16 '18 at 21:22