What is the inverse Laplace Transformation for $\exp (a s^4)$

Question

I need to find the inverse Laplace transform for $$ \exp(a\cdot s^4) $$ where a is a constant.

Alternatively, I need to calculate the following integral: $$f(t)= \frac{1}{2\pi i}\int_{\alpha -i\infty }^{\alpha +i\infty }e^{ as^{4}+ts} ds $$

Answer 1

Use Taylor expansion as follows: \begin{equation*} e^{as^{4}}=\sum\limits_{n=0}^{\infty }\frac{a^{n}}{n!}s^{4n} \end{equation*}

Apply inverse Laplace transform for two sides, then \begin{equation*} L^{-1}\left( \mathbf{e}^{as^{4}}\right) =\sum\limits_{n=0}^{\infty }\frac{a^{n}}{n!}L^{-1}\left( s^{4n}\right) \end{equation*} use the following relation \begin{equation*} L^{-1}\left( s^{n-1}\right) =\frac{d^{n-1}}{dt^{n-1}}\delta \left( t\right) ,\qquad n=1,2,... \end{equation*} use \begin{equation*} n=4k+1,\qquad k=0,\frac{1}{4},\frac{1}{2} \end{equation*} therefore, we have \begin{equation*} L^{-1}\left( \mathbf{e}^{as^{4}}\right) =\sum\limits_{k=0}^{\infty }\frac{a^{4k+1}}{\left( 4k+1\right) !}\frac{d^{4k}% }{dt^{4k}}\delta \left( t\right) ,\qquad k=0,\frac{1}{4},\frac{1}{2} \end{equation*}