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!
 A: The answer is :
$$S(a)=\sum_{k\geq 0}\frac{\left( -1\right) ^{k}\left( 2k+1\right) }{\left(
2k+1\right) ^{2}+a^{2}}=\frac {\pi}{4\,\cosh(\pi a/2)}$$
Let's start nearly as you did :
\begin{eqnarray*}
S(a)=\sum_{k\geq 0}\frac{\left( -1\right) ^{k}\left( 2k+1\right) }{\left(
2k+1\right) ^{2}+a^{2}} &=&\frac 12\sum_{k\geq 0}\left( -1\right) ^{k}%
\left[ \frac 1{2k+1+ia}-\frac 1{-2k-1+ia}\right]  \\
&=&\frac 12%
\sum_{k=-\infty}^\infty \frac{\left( -1\right) ^{k}}{2k+1+ia} \\
&=&\frac 14%
\sum_{k=-\infty}^\infty \frac{\left( -1\right) ^{k}}{k+(1+ia)/2} \\
\end{eqnarray*}
But the last series was evaluated earlier here using different methods with the result (in our case) :
$$S(a)=\frac {\pi}{4\sin\left(\pi (1+ia)/2\right)}=\frac {\pi}{4\cos\left(\pi a i/2\right)}=\frac {\pi}{4\cosh\left(\pi a/2\right)}$$

Since your method is fine let's try to finish it your way :
$$
\begin{align}
S(a)&=\frac 12\int_{0}^{1}\frac{x^{ia}+x^{-ia}}{1+x^{2}}dx\\
&=\int_{0}^{1}\frac{\cos(a\ln(x))}{1+x^{2}}dx\quad\text{using the hint in the comment}\\
&=\int_{-\infty}^0\frac{\cos(a\,t)\,e^t}{1+e^{2t}}dt\quad\text{if}\ \ x:=e^t\\
&=\frac 12\int_{-\infty}^0\frac{\cos(a\,t)}{\cosh(t)}dt\\
&=\frac 12\int_0^\infty \frac{\cos(a\,t)}{\cosh(t)}dt\\
\end{align}
$$
Let's evaluate $\displaystyle I(a)=\int_{-R}^R \frac{e^{i a\,z}}{\cosh(z)}dz\ $ using a rectangular contour $[-R,+R]\times [0,\pi i]\ $ (we will have $R\to\infty$ and suppose $|a|<1$) 
$$
I(a)=\int_{-R}^R \frac{e^{i a\,z}}{\cosh(z)}dz+\int_R^{R+\pi i} \frac{e^{i a\,z}}{\cosh(z)}dz+\int_{R+\pi i}^{-R+\pi i}\frac{e^{i a\,z}}{\cosh(z)}dz+\int_{-R+\pi i}^{-R} \frac{e^{i a\,z}}{\cosh(z)}dz
$$
putting the third integral in second position :
$$
I(a)=\int_{-R}^R \frac{e^{i a\,t}}{\cosh(t)}dt+\int_{R}^{-R}\frac{e^{i a(t+\pi i)}}{\cosh(t+\pi i)}dt+\int_R^{R+\pi i} \frac{e^{i a\,z}}{\cosh(z)}dz+\int_{-R+\pi i}^{-R} \frac{e^{i a\,z}}{\cosh(z)}dz
$$
I'll let you show that the two integrals at the right disappear as $R\to\infty$ (for $|a|<1$) and use $\cosh(t+\pi i)=-\cosh(t)$ to get :
$$I(a)=\int_{-\infty}^\infty \frac{e^{i a\,t}}{\cosh(t)}dt+\int_{-\infty}^\infty\frac{e^{i a(t+\pi i)}}{\cosh(t)}dt=\left(1+e^{-a\pi}\right)\int_{-\infty}^\infty \frac{e^{i a\,t}}{\cosh(t)}dt$$
Since there is only one pole at $\ t=\frac {\pi}2 i\ $ we may use the residu there to get :
$$\frac{2\pi i\, e^{i a\,\pi i/2}/i}{\left(1+e^{-\pi a}\right)}=\frac{2\pi\, e^{-\pi a/2}}{\left(1+e^{-\pi a}\right)}=\frac{\pi}{\cosh(\pi a/2)}=\int_{-\infty}^\infty \frac{e^{i a\,t}}{\cosh(t)}dt$$
and conclude that indeed :
$$\frac 12\int_0^\infty \frac{\cos(a\,t)}{\cosh(t)}dt=\frac{\pi}{4\cosh(\pi a/2)}$$
Perhaps not as simple as you hoped...
A: 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
$$ S(a) = \frac{1}{4} \frac{\pi}{\cosh \left(\frac{\pi a}{2}\right)}.$$
A: The series is summable and has the closed form formula
$$-\frac{1}{4}\,{\frac {\pi }{\sin \left( \frac{1}{2}\pi \, \left( 3 + ia \right ) 
\right) }} = \frac{1}{4}{\frac {\pi }{\cosh \left( \frac{\pi \,a}{2} \right) }} .$$
