Poisson Summation and Odd Functions Poisson Summation formula below works for both even and odd functions.
$$\sum\limits_{n=-\infty}^{\infty} f(n) = \sum\limits_{k=-\infty}^{\infty}\hat{f}(k)$$
Unlike even functions, this formula doesn't doesn't seem to be as useful for ODD functions, since both LHS and RHS of the identity compute to $0$. 
Question 1: Is there a one sided version that relates a sampled sum of a function with its FT( i.e. relating $\sum\limits_{n=1}^{\infty} f(n)$ to its frequency components)?

I found a variant of Poisson's sum that is defined using cosine transform in the 1948 book "Introduction to the theory of Fourier Integrals" by Titchmarsh E.C.
$$\sqrt{\beta}\Big(\frac{1}{2}F_c(0) + \sum\limits_{n=1}^{\infty}F_c(n\beta)\Big) = \sqrt{\alpha}\Big(\frac{1}{2}f(0) + \sum\limits_{n=1}^{\infty}f(n\alpha)\Big), \alpha\beta = 2\pi, \alpha > 0$$
For this formula, $F_c(x) = \sqrt{\frac{2}{\pi}}\int\limits_{0}^{\infty}f(t)cos(xt) dt $ for $x > 0 $
Question 2 : I am unsure if I can use this formula for odd functions. Double-sided cosine transform of an odd function results in 0, but since the formula in the book for cosine transform is one-sided, I am tempted to use it for the purpose stated in my Question 1 above. Any thoughts on this?
 A: Maybe there is another answer, but let me introduce the (tempered) distribution point of view.
The definition of the Fourier transform of a (tempered) distribution $T$ is
$$\langle T,\varphi = \langle \hat{T},\hat{\varphi}\rangle$$
for any test function $\varphi$,
that is to say the Fourier transform preserves the "inner product" (it is a unitary linear operator).
In that framework, the Poisson summation formula simply says that the Dirac comb distribution $\Pi\text{I}(x) = \sum_{n=-\infty}^\infty \delta(x-n)$ is its own Fourier transform : $\ \Pi\text{I} = \hat{\Pi\text{I}}$ and
$$\sum_{n=-\infty}^\infty \varphi(n) = \langle \Pi\text{I},\varphi \rangle = \langle \hat{\Pi\text{I}},\hat{\varphi} \rangle = \langle \Pi\text{I},\hat{\varphi} \rangle  = \sum_{n=-\infty}^\infty \hat{\varphi}(n)$$
(you already know that from the Fourier series, where  $\hat{\Pi\text{I}}(x) = \sum_{n=-\infty}^\infty e^{-2i \pi n x} =\delta_{\scriptstyle(1\text{ periodic})}(x) $)
Now, the Fourier transform of the one-sided Dirac comb $\Pi(x) = \sum_{n=1}^\infty \delta(x-n)$ is $$\hat{\Pi}(\xi) = \sum_{n=1}^\infty e^{-2i\pi n \xi} = \frac{1}{e^{2i \pi \xi}-1}$$
(the series converging in the sense of distributions)
So that
$$\sum_{n=1}^\infty \varphi(n) = \langle \Pi, \varphi \rangle = \langle \hat{\Pi}, \hat{\varphi} \rangle = \int_{-\infty}^\infty \frac{\hat{\varphi}(\xi)}{e^{-2i \pi \xi}-1}d\xi$$
And there should be no major simplification of this in general.
