# Integral of the ratio of two exponential sums

I am trying to find a lower bound on the following integral

\begin{align*}
\int_{y=-\infty}^{y=\infty} \frac{ (\sum_{n=[-N..N]/\{0\}}n e^{-\frac{(y-cn)^2}{2}})^2} {\sum_{n=[-N..N]/\{0\}} e^{-\frac{(y-cn)^2}{2}}}dy
\end{align*}

I have been trying to find the upper and the lower bounds on the summations in here.

I have also look at the Jacobian-Theta functions but I wasn’t able to find theta function that would match my cases.

Another approach is to look at hyperbolic functions
\begin{align*}
&\int_{y=-\infty}^{y=\infty} \frac{ (\sum_{n=[-N..N]/\{0\}}n e^{-\frac{(y-cn)^2}{2}})^2} {\sum_{n=[-N..N]/\{0\}} e^{-\frac{(y-cn)^2}{2}}}dy\\
&=\int_{y=-\infty}^{y=\infty} \frac{ e^{-y^2/2}\left(\sum_{n=[1..N]}n e^{-\frac{c^2n^2}{2}} \sinh(ncy)\right)^2} {\sum_{n=[1..N]} e^{-\frac{c^2n^2}{2}} \cosh(ncy)}dy\\
\end{align*}

But how to compute or bound the hyperbolic sums? Again, I only require a lower bound on the above integral.

Should elliptic functions come into the play here, but I am not sure?
I feel like the sums inside of the integral were already looked at, so I would also be very grateful if you can point me to some references.

Please see a very nice approach by @Dr. MV. The point of starting a bounty is to maybe show a tighter result.

#### Solutions Collecting From Web of "Integral of the ratio of two exponential sums"

We seek a lower bound for the integral $I_N$ where

$$I_N=\int_{-\infty}^{\infty} \frac{ (\sum_{1\le |n|\le N}n e^{-(y-cn)^2/2})^2} {\sum_{1\le |n|\le N} e^{-(y-cn)^2/2}}dy$$

Let the integrand be denoted by the square of the function $f_N$ such that

$$f_N^2(x)=\frac{ (\sum_{1\le |n|\le N}n e^{-(y-cn)^2/2})^2} {\sum_{1\le |n|\le N} e^{-(y-cn)^2/2}}$$

Also, let the denominator of $f_N^2$ be denoted by the square of the function $g$ such that

$$g_N^2(x)=\sum_{1\le |n|\le N} e^{-(y-cn)^2/2}$$

Applying the Cauchy-Schwarz inequality

$$\left(\int_{-\infty}^{\infty} f_N(x)g_N(x)dx\right)^2\le \int_{-\infty}^{\infty} f_N(x)^2dx\,\,\int_{-\infty}^{\infty} g_N^2(x)dx$$

implies that

\begin{align} I_N&=\int_{-\infty}^\infty f_N(x)^2 \, dx\\\\ &\ge \frac{\left(\int_{-\infty}^\infty f_N(x)g_N(x)\,dx\right)^2}{\int_{-\infty}^\infty g_N^2(x)\,dx}\\\\ &=\frac{\left(\int_{-\infty}^\infty \left|\sum_{1\le |n|\le N} n e^{-(y-cn)^2/2}\right|\,dx\right)^2}{\int_{-\infty}^\infty \sum_{1\le |n|\le N} e^{-(y-cn)^2/2}\,dx}\\\\ &=2\sqrt{\pi}\frac1 N \left(\sum_{n=1}^N n\operatorname{erf}(cn)\right)^2 \end{align}