# Thread: laplace transforms. singularities etc.

1. ## laplace transforms. singularities etc.

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

2. Originally Posted by walleye
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

[snip]
$\displaystyle s = 0$ is a removable singularity of $\displaystyle F(s) = \frac{\ln(s^2 +1)}{s}$ if $\displaystyle \lim_{s \rightarrow 0} \frac{\ln(s^2 +1)}{s}$ exists and is finite. To find this limit, use l'Hopital's Rule (since it's an indeterminant form 0/0):

$\displaystyle \lim_{s \rightarrow 0} \frac{\ln(s^2 +1)}{s} = \lim_{s \rightarrow 0} \frac{\frac{2s}{s^2 + 1}}{1} = 0$.

So yes s = 0 IS a removable singularity of F(s), not a simple pole. (The fact that the Laurent series has no principle part is also a big clue ....)