In a mathematical physical problem, the sum below needs to be calculated:
$$ F_n(\xi) = \sum_{k=0}^{\operatorname{ceil} \left(\frac{n}{2}-1 \right)} (-1)^k \frac{\xi^{2k}}{n-2k} = \begin{cases} \qquad\qquad\qquad\qquad\qquad\qquad ? & \text{if $n$ is even} \, , \\ (-1)^{\frac{n-1}{2}} \xi^{n-1} \Big( 1+\frac{1}{2} \Phi \left( -\xi^2,1,-\frac{1}{2} \right) \Big) - \frac{1}{2} \Phi \left( -\xi^2, 1, -\frac{n}{2} \right) & \text{if $n$ is odd} \, , \end{cases} $$ wherein $\xi$ is a positive real number. Here, $\operatorname{ceil}(x)$ denotes the ceiling function, which maps a variable $x$ to the least integer greater than or equal to $x$. Accordingly, $$ \operatorname{ceil}\left( \frac{n}{2}-1 \right)= \begin{cases} \frac{n}{2}-1 & \text{if $n$ is even} \, , \\ \frac{n-1}{2} & \text{if $n$ is odd} \, . \end{cases} $$
When $n$ is an odd number, it turned out that the sum can conveniently be expressed in terms of the Lerch transcendent function. However, when $n$ is even, i have no idea how to evaluate the sum.
It would be highly appreciated if someone here could try to be of help and let me know whether an analytical evaluation is possible. Any hint would be most welcome!
If $n=2m$, Maple says the sum is $$ -{\frac { \left( -{\xi}^{2} \right) ^{m}}{2\;m} \left( \ln \left( { \frac {{\xi}^{2}+1}{{\xi}^{2}}} \right) m+ \left( {\xi}^{2} \right) ^{ -m} \left( -1 \right) ^{m} \left( {\it LerchPhi} \left( -{\xi}^{-2},1, m \right) m-1 \right) \right) } $$