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What is the inverse Laplace transform of $$ \bar{U}(r,s) = \frac{1}{s} \frac{K_0 \left(\sqrt{s/\alpha} \, r \right)}{K_0 \left(\sqrt{s/\alpha} \, b \right)} \tag{12}. $$ ?

Related (but not providing the answer):

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    Note that you can assume that $\alpha=r=1$. – Gary Jan 25 '24 at 00:19
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    The solution to this particular inverse Laplace transform for the same problem is found in "Conduction of Heat in Solids" by H.S. Carslaw and J.C. Jaeger (Second Edition) - Clarendon Press. Ch. XIII p. 335. The inverse Laplace transform has a branch point at the origin and the integral is evaluated on each side of the branch cut on the negative axis. The final result is provided as an integral of a combination of standard Bessel functions $J_{0}$ and $Y_{0}$. – Jean-Luc Boulnois Jan 25 '24 at 02:11
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    Also, in this book, the specifics of the contour integration are discussed, without proof, in the section "Inversion Theorem for the Laplace Transformation" on p. 302, 303, as well as in Appendix $I$ p.479. – Jean-Luc Boulnois Jan 25 '24 at 02:27
  • @Jean-LucBoulnois - If you post that as an answer, I would accept it. – sancho.s ReinstateMonicaCellio Jan 25 '24 at 09:29
  • @Gary - Yes, I had just found it. If I am not wrong, it contains essentially the same information as the reference by Jean-Luc Boulnois. Did you check your solution is the same as in Carslaw & Jaeger, including signs, factors? – sancho.s ReinstateMonicaCellio Jan 25 '24 at 11:38

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