Use Parseval's equation and the table of Fourier series to evaluate $\sum\frac{1}{(1+n^2)^2}$ 
Use Parseval's equation and the table of Fourier series to evaluate $$\sum\frac{1}{(1+n^2)^2}$$

So I have used this method before to show $$\sum\frac{1}{n^4} = \frac{\pi^4}{32}$$ however, for this question I am struggling.
Basically, off the sheet I have from university I have used $$F(x) = \left|\sin\left(\pi\times\frac xL\right)\right| 
, -L<x<L, \to \frac 2\pi - \frac 4\pi \sum \frac{1}{n^2-1}\times\cos\left(\frac{n\pi x}{L}\right)$$
I find $a_0$ and $a_n$ then
I apply Parseval's theorem then it becomes clear I have gone wrong...
 A: Instead of Parseval's identity, we may just use differentiation under the integral sign.
By integration by parts we have, for any $a>0$ and $n\in\mathbb{N}^+$:
$$ \frac{1}{a^2+n^2} = \int_{0}^{+\infty}\frac{\sin(nx)}{n} e^{-ax}\,dx \tag{1}$$
hence by differentiating both sides with respect to $a$ we get:
$$ \frac{a}{(a^2+n^2)^2} = \frac{1}{2}\int_{0}^{+\infty}\frac{\sin(nx)}{n} x e^{-ax}\,dx \tag{2}$$
and by evaluating at $a=1$, then summing over $n\geq 1$, we get:
$$ \sum_{n\geq 1}\frac{1}{(n^2+1)^2} = \frac{1}{2}\int_{0}^{+\infty}W(x)\,x\, e^{-x}\,dx \tag{3} $$
where $W(x)$ is a sawtooth wave, a $2\pi$-periodic function that equals $\frac{\pi-x}{2}$ on the interval $(0,2\pi)$.
By exploiting such periodicity we get that:
$$ \sum_{n\geq 1}\frac{1}{(n^2+1)^2}=\frac{1}{2}\int_{0}^{2\pi}\frac{\pi-x}{2}\sum_{m\geq 0}(x+2m\pi)e^{-x-2m\pi}\,dx \tag{4}$$
and by simplifying the RHS we get:
$$ \sum_{n\geq 1}\frac{1}{(n^2+1)^2}=\frac{1}{4\left(e^\pi-e^{-\pi}\right)^2}\int_{0}^{2\pi}(\pi -x)e^{-x}(2\pi-x+ x e^{2\pi})\,dx \tag{5} $$
with the original problem boiling down to the evaluation of an elementary integral. The final outcome is:
$$ \boxed{\sum_{n\geq 1}\frac{1}{(n^2+1)^2} = \color{red}{-\frac{1}{2}+\frac{\pi\cosh(\pi)}{4\sinh(\pi)}+\frac{\pi^2}{4\sinh^2\pi}}}\tag{6}$$
A: It is easy to show that the complex Fourier Series for $e^{bx}$, $x\in[0,2\pi]$ is given by
$$e^{bx}=\frac{e^{2\pi b}-1}{2\pi}\sum_{n=-\infty}^\infty \frac{e^{inx}}{b-in}$$
whereupon isolating the series becomes
$$\sum_{n=-\infty}^\infty \frac{e^{inx}}{b-in}=\frac{2\pi e^{bx}}{e^{2\pi b}-1} \tag1$$
Differentiating $(1)$ with respect to $b$ and multiplying by $-1$ reveals
$$\begin{align}
\sum_{n=-\infty}^\infty \frac{e^{inx}}{(b-in)^2}&=-\frac{d}{db}\left(\frac{2\pi e^{bx}}{e^{2\pi b}-1} \right)\\\\
&=\left(\frac{2\pi e^{bx}}{(e^{2\pi b}-1)^2}\right)\left(2\pi e^{2\pi b}-(e^{2\pi b}-1)x \right)\tag 2
\end{align}$$
Next, setting $b=1$ in $(2)$ yields
$$\begin{align}
\sum_{n=-\infty}^\infty \frac{e^{inx}}{(1-in)^2}&=\left(\frac{2\pi e^{x}}{(e^{2\pi }-1)^2}\right)\left(2\pi e^{2\pi }-(e^{2\pi }-1)x \right)\tag 3
\end{align}$$
Finally, applying Parseval's Theorem to $(3)$ we find that 
$$\begin{align}
\sum_{n=-\infty}^\infty \frac{1}{(n^2+1)^2}&=\left(\frac{2\pi}{(e^{2\pi}-1)^2}\right)^2\,\frac{1}{2\pi}\int_0^{2\pi }e^{2x}(2\pi e^{2\pi}-(e^{2\pi}-1)x)^2\,dx\\\\
&=\frac{\pi}2 \left(\pi \text{csch}^2(\pi)+\text{coth}(\pi)\right)
\end{align}$$
whereupon solving for the series of interest yields
$$\bbox[5px,border:2px solid #C0A000]{\sum_{n=1}^\infty \frac{1}{(n^2+1)^2}=-\frac12 +\frac{\pi}4 \left(\pi \text{csch}^2(\pi)+\text{coth}(\pi)\right)}$$
