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    As part of a question to verify that \mathcal{L} (t^{\frac{1}{2}}) = \sqrt{\frac{\pi}{s}} , I have to use complex contour integration with the Bromwich contour to evaluate the inverse laplace transform of \sqrt{\frac{\pi}{s}}. We obviously know that the answer is t^{0.5} but I'm not sure how to work it out. I don't know what contour to take...

    I've tried taking the contour which goes like... \{z=iy : y\leq \epsilon\}\cup \{z=\epsilon e^{i*\theta} : \frac{\pi}{2}\leq\theta\leq \frac{3\pi}{2}\} \cup \{z=iy : y\geq \epsilon\} and then considering the limit as \epsilon \rightarrow 0 but this just gives me 0 which is wrong. So I'm confused as to whether I've taken the wrong contour or have calculated this incorectly.

    The hint says to take a contour similar to a keyhole...but alas I still cannot think of what it should be. Any help would be greatly appreciated - thanks in advance

    Because sqrt(z) has a branch point, you need to put in a cut line so that sqrt(z) will be single valued on the region you're integrating.

    As I understand it, (and as well as I can describe it on here), take a cutline along the -ve x (real) axis.

    you then want a contour going from 1-iR to 1+iR, then a quarter semicircle radius R ending at (1-R, 0), then a straight line from (1-R) to (-eps, 0) (above the cutline), then a clockwise circle radius eps going to (-eps, 0) (below the cutline), then a straight line from (-eps, 0) to (1-R), and finally a quarter semicircle radius R going from (1-R, 0) to (1- iR) again to close the contour. Then let R go to infinity.

    As I understand it, the two straight line integrals (from x=1-R to eps and back again) do NOT cancel out because sqrt(z) has a different value depending on whether you are above or below the cut.

    Not sure how difficult this will be to actually do - it's been too long since I did Contour Integration in anger - but good luck!
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Updated: April 14, 2011
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