the inversion laplace transformation of

$\displaystyle \frac{ln(s^2 +1)}{s}$

using the complex inversion formula

i dont understand why s=0 isnt considered a removable singularity, since

s=0 within ln(s^2 + 1) = ln (1) = 0

the lecturer's solutions give s=0 as a simple pole

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second question (unrelated to the above problem) concering complex inversion formula problems, with bromich contours

just for verification

considering a function that has 2 branch points (finite points) and a branch cut between the 2 (the cut not extending to infinity) the bromich contour is able to enclose the two branch points and the branch cut, provided you have a second contour scaling just beside and branch cut and around the branch points.

is this correct? do you understand my description?

thankyou nonetheless