Approximating an infinite sum of only odd terms by a definite integral Consider the infinite Sum
$S=\sum\limits_{n\ \text{odd}}^{\infty}\left(\dfrac{1}{nt}\right)^2\left[1-i\left(nt\right)^2\right]^{-1}$
Is there a way to approximate this sum as a contour integral? In my physical problem it is valid to make the approximation $t\to 0$ in a certain regime (or more precisely, the approximation that $t$ is very small). So I tried the following:
Let us set aside the fact (for now) that this is a sum over odd terms only. So we rewrite the sum as
$S=\dfrac{1}{t}\sum\limits_{n=1}^{\infty}t\cdot\left(\dfrac{1}{nt}\right)^2\left[1-i\left(nt\right)^2\right]^{-1}$
$\ \ =\dfrac{1}{t}\lim\limits_{t\to 0}\sum\limits_{n=1}^{\infty}t\cdot\left(\dfrac{1}{nt}\right)^2\left[1-i\left(nt\right)^2\right]^{-1}$
$\ \ =\dfrac{1}{t}\int\limits_0^\infty\dfrac{1}{x^2}\cdot\dfrac{1}{1-ix^2}\ dx$
(i.e.) setting $t\to dx$ and $nt\to x$ as this resembles a Riemann sum. But this doesn't work because the integral does not converge. Is there a nice way to approximate this infinite sum? Ideally, it'll be great if there is a way to do this without assuming $t$ is small.
EDIT: Corrected the problem statement based on Blumenthal's comment. 
 A: (This is an answer to a previous version of the question.)
You can express your sum as a power series in $t$.  First rewrite your sum as
$$
S = i \sum_{n = 0}^{\infty} \frac{\left(\frac{t}{2n+1}\right)^4}{1+i\left(\frac{t}{2n+1}\right)^2}.
$$
For $|t| < 1$ and $n\geq 0$ we have
$$
\frac{1}{1+i\left(\frac{t}{2n+1}\right)^2} = \sum_{m=0}^{\infty} (-i)^m \left(\frac{t}{2n+1}\right)^{2m},
$$
so that, by changing the order of summation,
$$
\begin{align}
S &= i \sum_{n = 0}^{\infty} \left(\frac{t}{2n+1}\right)^4 \sum_{m=0}^{\infty} (-i)^m \left(\frac{t}{2n+1}\right)^{2m+4} \\
&= i \sum_{m=0}^{\infty} (-i)^m t^{2m+4} \sum_{n=0}^{\infty} \frac{1}{(2n+1)^{2m+4}} \\
&= i \sum_{m=0}^{\infty} (-i)^m (1-4^{-m-2}) \zeta(2m+4) t^{2m+4},
\end{align}
$$
where we have used the identity
$$
\sum_{n=0}^{\infty} \frac{1}{(2n+1)^\alpha} = (1-2^{-\alpha})\zeta(\alpha).
$$
It follows, for example, that
$$
S = i \frac{\pi^4}{96} t^4 + \frac{\pi^6}{960} t^6 + O(t^8)
$$
as $t \to 0$.
A: You can have the following closed form for the sum
$$  -\frac{\sqrt{2}\left( 1+\,i \right) \pi}{32\,t^2} \, \left( (i-1)\pi \,\sqrt {2}+4\,t\,\tan
 \left( {\frac { \left( 1-i \right) \pi \,\sqrt {2}}{4\,t}}
 \right)  \right). $$
Check Here is the original answer by Maple in case I made a mistake while trying to improve the format in the above answer
$$ \left( -1/32-1/32\,i \right) \pi \, \left( i\pi \,\sqrt {2}+4\,\tan
 \left( {\frac { \left( 1/4-1/4\,i \right) \pi \,\sqrt {2}}{t}}
 \right) t-\pi \,\sqrt {2} \right) \sqrt {2}{t}^{-2}.$$
