Asymptotics of a series involving cos integral functions 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.




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


*

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

*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}

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:
$$ 2\sum_{n\in\mathbb{Z}}'\int_{n}^{n+y}\frac{dv}{v}=2\sum_{n\in\mathbb{Z}}'\log\left(1+\frac{y}{n}\right)=2\log\prod_{n\geq 1}\left(1-\frac{y^2}{n^2}\right)=\color{red}{2\log\left(\frac{\sin(\pi y)}{\pi y}\right)}.$$
