Inverse Laplace Transformation of an exponential function

Question

How one could find the inverse Laplace transformation of $\exp(-(b/(b+s))^k)$? Where both $b$ and $k$ are positive.

Answer 1

So you need to compute $$ \frac1{2\pi i}\int_{-i\infty}^{i\infty}e^{xs-\left(\tfrac{b}{b+s}\right)^k}ds=e^{-bx} \frac1{2\pi i}\int_{-i\infty}^{i\infty}e^{x(b+s)-\left(\tfrac{b}{b+s}\right)^k}ds. $$ For $x$ positive and $k$ positive integer this is equal to $$ e^{-bx}\operatorname{res}_{z=0}e^{xz-\left(\tfrac{b}{z}\right)^k}. $$ The residue is the coefficient of $z^{-1}$ in the Laurent series: $$ \operatorname{res}_{z=0}e^{xz-\left(\tfrac{b}{z}\right)^k}=\sum_{n-kl=1}\frac{x^n}{n!}\frac{(-b)^{kl}}{l!}=x\sum_{l=0}^\infty\frac{(-bx)^{kl}}{l!(kl+1)!} $$ Now Mathematica says that $$ \sum_{l=0}^\infty\frac{y^{l}}{l!(kl+1)!}={}_0F_k\left(;\frac2k,\frac3k,\dots,\frac{k+1}k;\frac{y}{k^k}\right), $$ and it remains to subsitute $y=(-bx)^k$.

Answer 2

For the special case $k=1$ you should have the answer

$$ {{\rm e}^{-bt}} \left( i\sqrt {{\frac {b}{t}}} {\rm I_{1} \left(\,2\,\sqrt {-bt}\right)}+\delta \left( t \right) \right) ,$$

where $I_n(x)$ is the modified Bessel function of the first kind and $\delta(x)$ is the dirac delta function.

Answer 3

$\mathcal{L}^{-1}_{s\to t}\left\{e^{-\left(\frac{b}{b+s}\right)^k}\right\}$

$=\mathcal{L}^{-1}_{s\to t}\left\{\sum\limits_{n=0}^\infty\dfrac{(-1)^nb^{kn}}{n!(b+s)^{kn}}\right\}$

$=\mathcal{L}^{-1}_{s\to t}\left\{1+\sum\limits_{n=1}^\infty\dfrac{(-1)^nb^{kn}}{n!(b+s)^{kn}}\right\}$

$=\delta(t)+\sum\limits_{n=1}^\infty\dfrac{(-1)^nb^{kn}t^{kn-1}e^{-bt}}{n!\Gamma(kn)}$