laplace transforms. singularities etc.

walleye · #1 $Old$ October 28th, 2009, 02:39 AM

the inversion laplace transformation of
$\frac{ln(s^2 +1)}{s}$
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

mr fantastic · #2 $Old$ October 28th, 2009, 04:27 AM

Quote:

Originally Posted by walleye $View Post$

the inversion laplace transformation of
$\frac{ln(s^2 +1)}{s}$
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]

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

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

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 ....)