I'm trying to minimize an error bound $$\frac{4 \pi \exp(\cosh(a))}{\exp(a N) -1},$$ where $N$ is the step size for the trapezoidal rule and $-a < Im < a, a > 0$ is a strip bound which may be adjusted for different $N$. Now the paper with this example states that one can show with calculus that the bound is minimized with a value of $a$ close to $a = \log(2N)$. I tried taking the derivative and setting it $0$; however this didn't yield any proper result.
How did they manage to find that $a = \log(2N)$ closely minimizes this bound?
Of no interest at all ! Just to show how good is the approximation.
@Gary explained why the approximation is valid.
You could even have something closer using a Taylor expansion of the derivative $$f'=\sinh (a)-\frac{n \,e^{a n}}{e^{a n}-1}$$ around $a=\log(2n)$. Limited to first order $$f'=\left(-\frac{2^n n^{n+1}}{2^n n^n-1}+n-\frac{1}{4 n}\right)+\left(\frac{2^n n^{n+2}}{\left(2^n n^n-1\right)^2}+n+\frac{1}{4 n}\right) (a-\log (2 n))+O\left((a-\log (2 n))^2\right)$$ and solving for $a$ gives $$a=\log(2n)+\frac{1}{1+\frac{4 \left(2^n \left(2^n n^n+n-3\right) n^n+2\right) n^2}{\left(2^n n^n-1\right) \left(2^n n^n+4 n^2-1\right)}}$$ For $n=1$, the correction term is $\frac{5}{13}$ dropping to $\frac{465}{4337}$ for $n=2$ and to $\frac{53965}{1733653}$ for $n=3$. For $n=10$, it is $0.0025$.