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I'm looking for the asymptotic expansion( or value ) of the following function \begin{equation} F[y,t] = \sideset{}{'}\sum_{n \in \mathbb{Z}}\text{Ci}\big[\frac{n^2}{t}\big] - \text{Ci}\big[\frac{(n+y)^2}{t}\big], \quad t \in [0, \infty), \,\, y \in [0, 1] \end{equation} where the sum is performed for all nonzero integers and $\text{Ci}$ is the cos integral function \begin{equation} \text{Ci}(x) = -\int_x^{\infty} \frac{\cos t}{t} dt \end{equation}

I plot the alternative form of the function at $y = \frac{1}{2}$ \begin{equation} F[y = \frac{1}{2}, t] = \sum_{n=1}^{\infty} 2\text{Ci}\big[\frac{n^2}{t}\big] - \text{Ci}\big[\frac{(n+0.5)^2}{t}\big] - \text{Ci}\big[\frac{(n-0.5)^2}{t}\big] \end{equation} using Mathematica.

truncated at 20 terms

truncated at 20 terms

truncated at 400 terms

truncated at 400 terms

The general pattern of these figures is that the function goes up and then has a damping process and finally converge to a fixed value at large $t$. My truncation is performed for $n \gg t$; it seems that beyond $n \gg t $, extra terms only increase the fuzziness of the curve, not the asympototic value.

I tried to approximate it using Euler-Maclaurin formula.

Let $f(n ) = 2\text{Ci}\big[\frac{n^2}{t}\big] - \text{Ci}\big[\frac{(n+y)^2}{t}\big] - \text{Ci}\big[\frac{(n-y)^2}{t}\big]$ then \begin{equation} F[y, t] = \lim_{N\rightarrow \infty} \int_1^{N} f(x) dx + \frac{1}{2}[ f(1) + f(N)] + \frac{1}{12}[f'(N) - f'(1)] \\- \frac{1}{720}[f^{(3)}(N) - f^{(3)}(1)] + \cdots \end{equation} Keeping up to the first derivative term, I have \begin{equation} F[y, t] \sim - 2\text{Ci}[\frac{1}{t}] + (\frac{1}{2} + y ) \text{Ci}[\frac{(1+y)^2}{t}] + (\frac{1}{2} - y ) \text{Ci}[\frac{(1-y)^2}{t}] \\- 2 \Big( \int_1^{1+y} + \int_1^{1-y} \Big) \cos \frac{u^2}{t}du \end{equation} However this function is very smooth at large $t$. It seems that I have to keep all the corrections at $n=1$ to have a reasonable estimation.

There are several questions I want to ask about the asymptotic behavior

  1. Does \begin{equation} \lim_{t\rightarrow \infty} F[y,t] \end{equation} exist?

  2. If the asymptotic value does exist, then is there some sort of series expansion available for the large $t$?

Thanks.


Edit:

Let me fill in the gaps in user Jack D'Aurizio's solution of asymptotic value(credit belongs to him).

First make a simple change of variable from $u$ to $v$: $u = \frac{v^2}{t}$ \begin{equation} -\sum_{n\in\mathbb{Z}}'\int_{n^2/t}^{(n+y)^2/t}\frac{\cos u}{u}\,du = -2\sum_{n\in\mathbb{Z}}'\int_{n}^{n+y}\cos(\frac{v^2}{t})\frac{dv}{v} \end{equation} The integral becomes an integration on $\mathbb{R}$ with standard measure and the integrand function \begin{equation} -2\cos(\frac{x^2}{t})\big[\sum_{n\in\mathbb{Z}}' \chi_{[n,n+y]}(x) \big] \end{equation} where $\chi(x)$ is the characteristic function.

Take $t$ to be an integer sequence, since \begin{equation} \Big|2\cos(\frac{x^2}{t})\big[ \sum_{n\in\mathbb{Z}}' \chi_{[n,n+y]}(x) \big]\Big| \le 2( \sum_{n\in\mathbb{Z}}' \chi_{[n,n+y]}(x) ) \end{equation} and the later is integrable( by using the Weierstrass product for the sine function ) \begin{equation} 2\sum_{n\in\mathbb{Z}}'\int_{n}^{n+y}\frac{dv}{v}=2\sum_{n\in\mathbb{Z}}'\log\left(1+\frac{y}{n}\right)=2\log\prod_{n\geq 1}\left(1-\frac{y^2}{n^2}\right)=2\log\left(\frac{\sin(\pi y)}{\pi y}\right). \end{equation} the dominated convergence theorem claims that \begin{equation} \lim_{t\rightarrow \infty} F[y,t] = -\int_{\mathbb{R}} dx\, \lim_{t\rightarrow \infty} 2\cos(\frac{x^2}{t})\big[ \sum_{n\in\mathbb{Z}}' \chi_{[n,n+y]}(x) \big] = -2\log\left(\frac{\sin(\pi y)}{\pi y}\right) \end{equation}

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So we want to study the behaviour for a large $t$ of: $$ \sum_{n\in\mathbb{Z}}'\int_{n^2/t}^{(n+y)^2/t}\frac{\cos u}{u}\,du = 2\sum_{n\in\mathbb{Z}}'\int_{n}^{n+y}\cos(v^2/t)\frac{dv}{v}$$ that by the dominated convergence theorem and the Weierstrass product for the sine function approaches: $$ 2\sum_{n\in\mathbb{Z}}'\int_{n}^{n+y}\frac{dv}{v}=2\sum_{n\in\mathbb{Z}}'\log\left(1+\frac{y}{n}\right)=2\log\prod_{n\geq 1}\left(1-\frac{y^2}{n^2}\right)=\color{red}{2\log\left(\frac{\sin(\pi y)}{\pi y}\right)}.$$

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    $\begingroup$ Thanks @Jack D'Aurizio, impressive solution! In the figure it is about 0.91 for $y = 1/2$, indeed very close to -$2 \ln \Big( \frac{\sin{\pi y }}{\pi y} \Big)$. The opposite sign is due to a missing minus sign in the definition of cos integral function(I have edited the post). Do you have idea in doing a systematic expansion in terms of $1/t$ or other relevant small variables in large time? $\endgroup$ – anecdote Apr 8 '16 at 0:08
  • $\begingroup$ @anecdote: well, in principle the remaining part is given by $$2\sum_{n\in\mathbb{Z}}'\int_{n}^{n+y}\sin^2\left(\frac{v^2}{2t}\right)\frac{dv}{v}$$ that looks manageable through Fourier series. Interesting things happen for $y=1$. I will keep working on this problem. $\endgroup$ – Jack D'Aurizio Apr 8 '16 at 1:56
  • $\begingroup$ Hi @Jack D'Aurizio, in reflection, I have questions about the use of dominated convergence theorem. What absolute convergent function do you use to control the $\cos(v^2/t) /v$? The function $1/|v|$ is not integrable. Or you use the measure $dv / v = \text{sgn}(v) d \ln v $? Is it a valid measure(it can be negative)? $\endgroup$ – anecdote Apr 8 '16 at 19:37

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