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Math Help - laplace transforms. singularities etc.

  1. #1
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    laplace transforms. singularities etc.

    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

    ~~~~~~~~~~~~~~~~~~

    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
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  2. #2
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    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 ....)
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