Application of Poisson summation Formula to Casimir Effect I'm studying the Casimir Effect at finite temperature. To calculate the Helmoltz free energy in the canonical ensemble I need to sum a particular series. In some scientific papers it is suggested to use a modified Poisson formula over positive integers only that appears to be:
$$\frac{1}{2}F(0) + \sum_{n=1}^{\infty} F(n) = \pi \tilde{F}(0) + 2\pi \sum_{k=1}^{\infty} \tilde{F}(2\pi k)$$
where
$$\tilde{F}(k):= \frac{1}{\pi} \int_0^\infty F(t) \cos(kt)dt.$$
I can't understand how to demonstrate the validity of this expression starting from the standard summation formula, that is:
$$\sum_{n \in \mathbb{Z}} F(2\pi n) = \frac{1}{2\pi} \sum_{k \in \mathbb{Z}} \hat{F}(k)$$
where
$$\hat{F}(k):= \int_{-\infty}^\infty F(x) e^{-i k x}dx$$
is the standard Fourier transform. In fact this formula is deduced using tempered distributions so $F$ (that is a "test" function) has to have good property of convergence: this mean I can't simply define a non-continuous function null for $n<0$. 
In my particular case $F(x) = (ax)^2 \log(1-e^{-ax})$.
Sorry for my bad English ;)
 A: We start with
\begin{align}
 \frac{1}{2\pi} \sum_{k\in {\mathbb Z}} {\hat f}(k) = \sum_{n\in{\mathbb Z}} f(2\pi n)
\end{align}
A little rewriting gives
\begin{align}
\frac{1}{2} {\hat f}(0) + \frac{1}{2} \sum_{k=1}^\infty \left[ {\hat f}(k) + {\hat f}(-k) \right] = \pi f(0) + \pi \sum_{n=1}^\infty \left[ f(2\pi n) + f(-2\pi n) \right] 
\end{align}
To reproduce the formula you are interested, we now substitute
$$
f(x) = {\tilde F}(x) = \frac{1}{\pi} \int_0^\infty F(t) \cos (x t ) dt
$$
This function satisfies $f(x) = f(-x)$. The Fourier transform of this function is
$$
{\hat f}(k) =  \int_0^\infty dt F(t) \left[ \delta(t-k) + \delta(t+k) \right] = F(k) \theta(k) + F(-k) \theta(-k)
$$
In particular, when $k>0$, we have ${\hat f}(k) = {\hat f}(-k) = F(k)$ and ${\hat f}(0) = F(0)$ (one gets an additional factor of 1/2 since when $k=0$, the integral is over half the real line). Putting all of this back into the equation, we find
\begin{align}
\frac{1}{2} F(0) +  \sum_{k=1}^\infty F(k) = \pi {\tilde F}(0) + 2 \pi \sum_{n=1}^\infty {\tilde F}(2\pi n)
\end{align}
QED. 
A: I) Special case $F(x)=F(-x)$ even: The modified Poisson resummation formula (OP's first formula) follows from the standard Poisson resummation formula 
$$\sum_{n \in \mathbb{Z}} F(n) ~=~  \sum_{k \in \mathbb{Z}} \hat{F}(2\pi k),\qquad  \hat{F}(k)~:=~ \int_{-\infty}^\infty F(x) e^{-i k x}dx,$$
via straightforward standard manipulations. 
II) General case: Note that the modified Poisson resummation formula does not refer to the negative $x$-axis (or $t$-axis) at all! Thus we may assume without loss of generality that $F$ is even $F(x)=F(-x)$, i.e. replace $F$ with $F\circ |\cdot|$. [Caveat: If $F$ is smooth to start with then $F\circ |\cdot|$ will in general not be smooth at $x=0$. However the standard Poisson resummation formula also holds for some appropriate class of non-smooth functions. See the Wikipedia page for further details.]
