# What is inverse Laplace of following function? F(s)=L^(-1) ((Gamma(−s/a+b+1/4))/(Gamma(1/4−s/a)))

What is inverse Laplace of following function?
$F\left(s\right)={L}^{-1}\left(\frac{\Gamma \left(-\frac{s}{a}+b+\frac{1}{4}\right)}{\Gamma \left(\frac{1}{4}-\frac{s}{a}\right)}\right)$
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rougertl
The ratio of the gamma functions grows as ${s}^{b}$, therefore we need $b<0$ in order for the Bromwich integral to be convergent. Further, we need all poles to lie to the left of a vertical line, which is possible if either $a<0$ or $b\in \mathbb{Z}$. In the latter case the poles in the numerator and in the denominator cancel out. With these assumptions,
${\mathcal{L}}^{-1}\phantom{\rule{negativethinmathspace}{0ex}}\left[\frac{\mathrm{\Gamma }\phantom{\rule{negativethinmathspace}{0ex}}\left(-\frac{s}{a}+b+\frac{1}{4}\right)}{\mathrm{\Gamma }\phantom{\rule{negativethinmathspace}{0ex}}\left(-\frac{s}{a}+\frac{1}{4}\right)}\right]=a{e}^{at\left(b+1/4\right)}\sum _{k\ge 0}\frac{\left(-1{\right)}^{k+1}{e}^{atk}}{\mathrm{\Gamma }\left(-k-b\right)k!}=\phantom{\rule{0ex}{0ex}}-\frac{a{e}^{at\left(b+1/4\right)}}{\mathrm{\Gamma }\left(-b\right)}\sum _{k\ge 0}\left(-1{\right)}^{k}\left(\genfrac{}{}{0}{}{-b-1}{k}\right){e}^{atk}=-\frac{a{e}^{at\left(b+1/4\right)}}{\mathrm{\Gamma }\left(-b\right)}\left(1-{e}^{at}{\right)}^{-b-1}.$