integral of sech(x) from -Inf to Inf using residues.
Calculate using: (2 Pi I) * Res[sech(x), "poles in upper half plane"]
The Attempt at a Solution
I used sech(x) = 2/[exp(x)+exp(-x)] to find a simple pole at z = (I Pi)/2 with a residue of -I. Then,
Result = (2 Pi I)(-I) = 2 Pi
As far as I know the answer should be Pi so I'm off by a factor of two. I can't find the mistake in finding the residue. Can anyone verify that the residue should be -I/2?
Let f(z)=g(z)/h(z), then residue of a simple pole z0 is g(z0)/h'(z0), no? (Assuming g(z0) != 0, h(z0)=0)