Since $\displaystyle \frac{1}{\sinh x}$ behaves like $\displaystyle \frac{1}{x}$ near $0$, the singularity at $x=0$ is removable.Let $ f(z) = \displaystyle \frac{1}{z^{2}} - \frac{1}{z \sinh z} $ and integrate around a rectangle with vertices at $z=N, z= N + i \pi (N+\frac{1}{2}),$
$ z = -N + i \pi (N+ \frac{1}{2}),$ and $z= - N,$ where $N$ is an positive integer.Letting $N$ go to infinty, $ \displaystyle \int \frac{dz}{z^{2}} $ and $ \int \displaystyle \frac{dz}{z \sinh z} $ will evaluate to zero along the top and sides of the rectangle.
But it's not obvious that $ \displaystyle \int \frac{dz}{z \sinh z}$ evaluates to $0$ along the top of the rectangle. So I'll show that.$ \displaystyle \Big| \int_{-N}^{N} \frac{dt}{[t+i \pi (N+\frac{1}{2})] \sinh[ t + i \pi(N+\frac{1}{2})]} \Big| \le \int_{-N}^{N} \frac{dt}{[\pi(N+\frac{1}{2})-t] \cosh t} $
$ \displaystyle \le \int_{-N}^{N} \frac{1}{\pi(N+\frac{1}{2}) \cosh t} \ dt \le \frac{1}{\pi(N+ \frac{1}{2})} \int_{-\infty}^{\infty} \frac{1}{\cosh t} \ dt = \frac{1}{N+\frac{1}{2}} \to 0$ as $N \to \infty$So $ \displaystyle \int_{0}^{\infty} \Big( \frac{1}{x^{2}} - \frac{1}{x \sinh x} \Big) \ dx = \frac{1}{2} \int_{-\infty}^{\infty} \Big( \frac{1}{x^{2}} - \frac{1}{x \sinh x} \Big) \ dx= \pi i \sum_{n=1}^{\infty} \text{Res} [f(z), n \pi i]$$\displaystyle \text{Res} [f,n \pi i] = \lim_{z \to n \pi i} \frac{\sinh z -z}{2z \sinh z + z^{2} \cosh x} = \frac{-n \pi i}{(n \pi i)^{2} (-1)^{n}} = (-1)^{n-1} \frac{1}{n \pi i}$$\implies \displaystyle \int_{0}^{\infty} \Big( \frac{1}{x^{2}} - \frac{1}{x \sinh x} \Big) \ dx = \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n} = \ln 2$
