The partial fraction decomposition of $\frac{x^{2m-1}}{\prod_{k=1}^{n} (x^{2}+k^{2})}$ How do you find the partial fraction decomposition of $$\frac{x^{2m-1}}{\prod_{k=1}^{n} (x^{2}+k^{2})} \, ,$$ where $m$ is some positive integer? 
I don't know how to approach this other than to find the decomposition for different values of $m$ and $n$ and try to notice a pattern.
Is there a more systematic way I can approach this?
 A: Let us modify the problem slightly and compute:
\begin{equation}
S^{(m)}_n(x) := \frac{x^{2 m}}{\prod\limits_{k=1}^n (x^2+k^2)}
\end{equation}
for $m < n$.
From the partial fraction decomposition (see Decomposition into partial fractions of an inverse of a generic polynomial with three distinct roots. for example) we immediately have:
\begin{equation}
S^{(0)}_n(x) = \frac{1}{(n-1)!} \sum\limits_{k=1}^n \binom{n-1}{k-1} \frac{(-1)^{k-1}}{(x^2+k^2)} \cdot \frac{k!}{(n+k)!} \cdot 2 k
\end{equation}
Now let us multiply both sides of the equation by $x^2$ and then in the fraction in the sum on the right hand side write $x^2 = (x^2+k^2)-k^2$. Dividing this by $(x^2+k^2)$ produces a constant. However due to the assumption $n<m$ we know that the lhs cannot have any constants in its partial fraction decomposition. Therefore the respective term on the rhs must disappear and we get:
\begin{equation}
S^{(2)}_n(x)=  \frac{1}{(n-1)!} \sum\limits_{k=1}^n \binom{n-1}{k-1} \frac{(-1)^{k-0}k^2}{(x^2+k^2)} \cdot \frac{k!}{(n+k)!} \cdot 2 k
\end{equation}
Repeating the procedure again it is already clear that we get:
\begin{equation}
S^{(m)}_n(x)=\frac{1}{(n-1)!} \sum\limits_{k=1}^n \binom{n-1}{k-1} \frac{(-1)^{k-1+m}k^{2 m}}{(x^2+k^2)} \cdot \frac{k!}{(n+k)!} \cdot 2 k
\end{equation}
for $1\le m < n$.
A: One standard trick is as follows. Write:
$$
\frac{x^{2 m}}{\prod_{1 \le k \le n}(x^2 + k^2)} 
   = \sum_{1 \le k \le n} \frac{a_k}{x - i k} + \frac{b_k}{x + i k}
$$
Multiply through by $x - i k$, and make $x \to \dfrac{1}{i k}$ (here $i = \sqrt{-1}$). On the right hand side only the term for $a_k$ survives. With a bit of luck you can get the value of the left hand side. Similarly for $x + i k$ and $b_k$.
