<s>[INST] 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!
 [/INST] 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
$$ S(a) = \frac{1}{4} \frac{\pi}{\cosh \left(\frac{\pi a}{2}\right)}.$$
</s>