My problem is to get the inverse Laplace transform of the following equation.
$$\hat{P}(s) = \frac{\Gamma(p+1+s T)}{p! N^{s T}}$$
$p$, $T$ and $N$ are positive constants.
The denominator $N^{-s T}$ can be transformed easily, and the result is $\delta(t – T ln(N))$. If the inverse form of $\Gamma(p+1+s T)$ is known, then the final result is the Convolution of two parts. I think this is not very hard for me.
I found Mellin's inverse formula, but I can not complete the calculation.
The final result is known, but I don't know how to get it.
$$p(x) \sim \frac{1}{p!}exp(-(p+1) x -e^{-x}) $$
where $x = t/T – ln(N)$
I hope some one could give me details in this derivation.
Best Answer
The Gamma function has poles at $z=0$ and at the negative integers. When $z=-k$, then the residue at that pole is $(-1)^k/k!$. Now, the inverse transform is
$$\frac1{i 2 \pi} \int_{c-i \infty}^{c+i \infty} ds \, \Gamma(p+1+T s) e^{s (t-T \log{N})} $$
The poles of this Gamma function are at $p+1+T s = -k$, or $s=-(p+1+k)/T$ for $k \in \{0,-1,-2,\ldots \}$. Thus, the ILT is
$$e^{-(p+1)(t-T \log{N}))/T} H(t-T \log{N})\sum_{k=0}^{\infty} \frac{(-1)^k}{k!} e^{-k (t-T \log{N})/T} $$
where $H$ is the Heaviside function, or, summing the series,
$$e^{-(p+1)(t-T \log{N}))/T} \exp{\left [-e^{-(t-T \log{N})/N}\right ]} H(t-T \log{N}) $$
which, aside from the Heaviside factor, agrees with your expected result.