sum of a series Can
\begin{equation}
\sum_{k\geq 0}\frac{\left( -1\right) ^{k}\left( 2k+1\right) }{\left(
2k+1\right) ^{2}+a^{2}},
\end{equation}
be summed explicitly, where $a$ is a constant real number? If $a=0,$ this
sum becomes
\begin{equation}
\sum_{k\geq 0}\frac{\left( -1\right) ^{k}}{2k+1}=\frac{\pi }{4}.
\end{equation}
What about for $a\neq 0$. I tried this method
\begin{eqnarray*}
\sum_{k\geq 0}\frac{\left( -1\right) ^{k}\left( 2k+1\right) }{\left(
2k+1\right) ^{2}+a^{2}} &=&\frac{1}{2}\sum_{k\geq 0}\left( -1\right) ^{k}%
\left[ \frac{1}{2k+1+ja}+\frac{1}{2k+1-ja}\right]  \\
&=&\frac{1}{2}\sum_{k\geq 0}\frac{\left( -1\right) ^{k}}{2k+1+ja}+\frac{1}{2}%
\sum_{k\geq 0}\frac{\left( -1\right) ^{k}}{2k+1-ja} \\
&=&\frac{1}{2}\sum_{k\geq 0}\left( -1\right) ^{k}\int_{0}^{1}x^{2k+ja}dx+%
\frac{1}{2}\sum_{k\geq 0}\left( -1\right) ^{k}\int_{0}^{1}x^{2k-ja}dx \\
&=&\frac{1}{2}\int_{0}^{1}\left[ \sum_{k\geq 0}\left( -1\right) ^{k}x^{2k+ja}%
\right] dx+\frac{1}{2}\int_{0}^{1}\left[ \sum_{k\geq 0}\left( -1\right)
^{k}x^{2k-ja}\right] dx \\
&=&\frac{1}{2}\int_{0}^{1}\frac{x^{ja}}{1+x^{2}}dx+\frac{1}{2}\int_{0}^{1}%
\frac{x^{-ja}}{1+x^{2}}dx \\
&=&\frac{1}{2}\int_{0}^{1}\frac{x^{ja}+x^{-ja}}{1+x^{2}}dx
\end{eqnarray*}
but I was stucked at the last equations. Can any one give me some hint or tell me that the analytic expression doesn't exist. Thanks very much!

This sum can be evaluated by the same trick as presented here:
math.stackexchange.com.
In order to capture the convergence behavior of the series group consecutive terms to obtain absolute convergence, writing
$$ S(a) = \sum_{m\ge 0} \frac{4m+1}{(4m+1)^2+a^2}
-  \sum_{m\ge 0} \frac{4m+3}{(4m+3)^2+a^2} =
\sum_{m\ge 0} 
\frac{(4m+1)((4m+3)^2+a^2)-(4m+3)((4m+1)^2+a^2)}
{((4m+3)^2+a^2)((4m+1)^2+a^2)} =
\sum_{m\ge 0}
\frac{2(4m+1)(4m+3)-2a^2}{((4m+3)^2+a^2)((4m+1)^2+a^2)}
$$
Instead of using $f_1(z)$ and $f_2(z)$ from the other post use
$$f(z) = 
\frac{2(4z+1)(4z+3)-2a^2}{((4z+3)^2+a^2)((4z+1)^2+a^2)} \pi \cot(\pi z).$$
The key operation of this technique is to compute the integral of $f(z)$ along a circle of radius $R$ in the complex plane, where $R$ goes to infinity. We certainly have $|\pi\cot(\pi z)|<2\pi$ for $R$ large enough. The core term from the sum is $\theta(R^2/R^4)$ which is $\theta(1/R^2)$ so that the integrals are $\theta(1/R)$ and vanish in the limit. This means that the sum of the residues at the poles of $f(z)$ add up to zero.
It is easily verified that the poles at the integers produce the terms of the sum twice over. It follows that
$$ 2 S(a)
+ \text{Res}_{z=\frac{3}{4}-\frac{1}{4}ia} f(z)
+ \text{Res}_{z=\frac{3}{4}+\frac{1}{4}ia} f(z)
+ \text{Res}_{z=\frac{1}{4}-\frac{1}{4}ia} f(z)
+ \text{Res}_{z=\frac{1}{4}+\frac{1}{4}ia} f(z)
= 0$$
The residues are easily computed as the poles are all simple.
This finally yields
$$ S(a) = \frac{1}{4} \frac{\pi}{\cosh \left(\frac{\pi a}{2}\right)}.$$
