# 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

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}

## 1 Answer

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

• 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? – anecdote Apr 8 '16 at 0:08
• @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. – Jack D'Aurizio Apr 8 '16 at 1:56
• 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)? – anecdote Apr 8 '16 at 19:37