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I am trying to prove
$$sum_{ngeq1}frac1{n^2+1}=frac{picothpi-1}2$$
Letting $$S=sum_{ngeq1}frac1{n^2+1}$$
we recall the Fourier series for the exponential function
$$e^x=frac{sinhpi}pi+frac{2sinhpi}pisum_{ngeq1}frac{(-1)^n}{n^2+1}(cos nx-nsin nx)$$
Plugging in $x=pi$
$$e^pi=frac{sinhpi}pi+frac{2sinhpi}pisum_{ngeq1}frac{(-1)^n}{n^2+1}(cos npi-nsin npi)$$
$$e^pi=frac{sinhpi}pi+frac{2sinhpi}pisum_{ngeq1}frac{(-1)^n}{n^2+1}((-1)^n-ncdot0)$$
$$e^pi=frac{sinhpi}pi+frac{2sinhpi}pi S$$
$$S=frac{pi e^pi}{2sinhpi}-frac12$$
But that is nowhere near to correct. What did I do wrong, and how do can I prove the identity? Thanks.

The formula
$$
e^x=frac{sinhpi}pi+frac{2sinhpi}pisum_{ngeq1}frac{(-1)^n}{n^2+1}(cos nx-nsin nx)
$$ is valid only for $|x|<pi$ since $e^x$ is regarded as $2pi$-periodic function extended from $(-pi, pi).$ Since $2pi$-periodic function $xmapsto e^x$ is of bounded variation, its Fourier series converges to the mean of its left and right limit at every point. So if you evaluate the RHS at $x=pi$, then you get
$$
frac{e^{pi}+e^{-pi}}{2} = coshpi = frac{sinhpi}pi+frac{2sinhpi}pi S,
$$ and hence
$$
S = sum_{ngeq1}frac1{n^2+1}=frac{picothpi-1}2,
$$as desired.

Recalling that $frac{psi(z)-psi(s)}{z-s}=sum_{nge 0} frac{1}{(n+z)(n+s)}$, your sum is equal to
$$sum_{nge 1} frac{1}{n^2+1}=frac{psi(i)-psi(-i)}{2i}-1$$
Now from two identities of the digamma function,
begin{align}
psi(1-z)-psi(z)&=picot(pi x) \
psi(1+z)-psi(z)&=frac{1}{z} \
end{align}
We may find that $psi(i)-psi(-i)=-frac{1}{i}-picot(pi i)$. Thus, we can conclude that$$sum_{nge 1} frac{1}{n^2+1}=frac{pi}{2}coth(pi)-frac{1}{2}$$