I'm trying to compute the inverse Laplace transform of $f(s) = s^c/(N + s^{ir} )$ where $c,N \in \mathbb{C}$ and $r \in \mathbb{R}^+$ using the Bromwich integral $$ F(t) = \frac{1}{2 \pi i} \int_{- i \infty + \epsilon}^{i \infty+ \epsilon}\,\, ds \frac{s^c e^{st} }{N + s^{ir} } $$ with the contour deformed around a branch cut along the negative real axis from infinity to the branch point $s=0$.

Mathematica will not compute it via the builtin InverseLaplaceTransform function.

I've tried to compute it by hand, decomposing the contour into a small circle $\{ |s|= \delta \ll 1 , -\pi < \theta \leq \pi\}$ around the branch point and the branch cut $\theta=\pi$ with $|s|$ running from $\delta$ to $\infty$. Mathematica can do these integrals if I express the exponential $e^{st}$ as an infinite sum, giving a rather intractable sum of hypergeometric functions ${}_2 F_{1}$.

If the integral cannot be done with standard analytic methods, it would be fine to do it numerically, but it would be really nice to have at least the scaling of $F$ for large $t$ analytically.

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    $\begingroup$ To get started, try solving your contour integral with c an integer. If you cannot do that, you will not be able to obtain the inverse transform of more complicated functions. In general, avoid using capital letters to begin the names of user-defined symbols. $\endgroup$ – bbgodfrey May 24 '16 at 0:56
  • $\begingroup$ It's the presence of the s^{ir} in the denominator causing the problems. The complexity of c is secondary I believe. $\endgroup$ – Peter Zimmerman May 24 '16 at 22:23
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    $\begingroup$ Hint: 1. Solution can be expressed by Mittag-Leffler function Hint: 2. Read about fractional derivatives (differential equations) and its integral transforms $\endgroup$ – user42761 Sep 1 '16 at 8:04
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    $\begingroup$ Answering old questions is fine, but I think OP is past the Hints stage by now. $\endgroup$ – Feyre Sep 1 '16 at 8:33

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