inverse Laplace of Gamma function

Question

what is inverse Laplace of following function? $$F(s)=L^{-1}(\frac{Γ(-\frac{s}{a}+b+\frac{1}{4})}{Γ(\frac{1}{4}-\frac{s}{a})})$$ I know this phrase has a pole in $s=a(b+\frac{1}{4}+k)$. $k=0,1,2,...$ therefore $Res(e^{st}F(s),-k)=\sum_{k=0}^{\infty}e^{a(b+\frac{1}{4}+k)t}\frac{Γ(-k)}{Γ(-b-k)}=\sum_{k=0}^{\infty}e^{a(b+\frac{1}{4}+k)t}\frac{(-1)^{k}}{k!Γ(-b-k)}$. The gamma function in the denominator can be written like this in the form of a numerator?Can anyone help me with this question? Thank you

Answer 1

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, $$\mathcal L^{-1} \!\left[ \frac {\Gamma \!\left( -\frac s a + b + \frac 1 4 \right)} {\Gamma \!\left( -\frac s a + \frac 1 4 \right)} \right] = a e^{a t (b + 1/4)} \sum_{k \geq 0} \frac {(-1)^{k + 1} e^{a t k}} {\Gamma(-k - b) k!} = \\ -\frac {a e^{a t (b + 1/4)}} {\Gamma(-b)} \sum_{k \geq 0} (-1)^k \binom {-b - 1} k e^{a t k} = -\frac {a e^{a t (b + 1/4)}} {\Gamma(-b)} (1 - e^{a t})^{-b - 1}.$$