I would like to compute the inverse Laplace of
$$F(s) = e^{-s^{\alpha}}$$
For the case $\alpha = \frac{1}{2}$ I've found the following on this website:
Compute the inverse Laplace transform of $e^{-\sqrt{z}}$
However, the accepted solution uses completing the square as one of its final steps, which is not possible fo $\alpha \neq \frac{1}{2}$, so I could no simply adjust the solution to solve my problem.
I come to the point where I have to solve:
$$f(t)=\int_{0}^{\infty}\frac{2}{\pi}ue^{u^2t}sin(u^{2\alpha}) du $$
to get the inverse function.
Any help is appreciated.
You are not going to integrate this in elementary terms.
One way to proceed, which requires some care in interpretationn is: $\mathrm{e}^{-s^\alpha} = \sum_{k=0}^\infty \frac{(-s^\alpha)^k}{k!}$, so \begin{align*} \mathscr{L}^{-1} \left\{ \mathrm{e}^{-s^\alpha} \right\}(t) &= \mathscr{L}^{-1} \left\{ \sum_{k=0}^\infty \frac{(-s^\alpha)^k}{k!} \right\}(t) \\ &\overset{?}{=} \sum_{k=0}^\infty \mathscr{L}^{-1} \left\{ \frac{(-s^\alpha)^k}{k!} \right\}(t) \\ &= \sum_{k=0}^\infty \frac{(-1)^k t^{-k\alpha - 1}}{k! \Gamma(-k \alpha)} \text{,} \end{align*} where the "?" is because we could easily have lost equality due to swapping limits. And we did: that denominator is trouble in the first term of the sum ... and that problem is replicated into every term if $\alpha$ is a nonnegative integer.
We can sidestep the problem in the first term. Let $$ f(\alpha) = \lim_{k \rightarrow 0} \left( \frac{(-1)^k t^{-k \alpha - 1}}{k! \Gamma(-k \alpha)} \right) + \sum_{k=1}^\infty \frac{(-1)^k t^{-k \alpha - 1}}{k! \Gamma(-k \alpha)} \text{,} $$ wherever that series converges. Then, where $\mathrm{Ai}(x)$ is the Airy function, $J_1(x)$ is the Bessel function of the first kind, and ${}_0\mathrm{F}_n(q_1, q_2, \dots, q_n; z)$ is a generalized hypergeometric function, \begin{align*} f(1/2) &= \frac{\mathrm{e}^{-1/(4t)}}{2 \sqrt{\pi}t^{3/2}} \\ f(1/3) &= 3^{-1/3}t^{-4/3}\mathrm{Ai}(3^{-1/3}t^{-1/3}) \\ f(2/3) &= \frac{2}{3^{4/3}}t^{-7/3}\mathrm{e}^{-2/(27t^2)} \left( \mathrm{Ai}(3^{-4/3}t^{-4/3}) - 3^{2/3} t^{2/3} \mathrm{Ai}'(3^{-4/3}t^{-4/3}) \right) \\ f(1/4) &= \frac{-{}_0\mathrm{F}_2(1/2, 3/4; -1/(256t))}{t^{5/4}\Gamma(-1/4)} - \frac{{}_0\mathrm{F}_2(3/4,5/4;-1/(256t))}{4 \sqrt{\pi}t^{3/2}} - \frac{{}_0\mathrm{F}_2(5/4,3/2;-1/(256t))}{6 t^{7/4}\Gamma(-3/4)} \\ f(-1/2) &= \frac{-{}_0\mathrm{F}_2(1/2, 3/2; t/4)}{\sqrt{\pi t}} + \frac{1}{2}{}_0\mathrm{F}_2(3/2,3; t/4) \\ f(-1) &= -J_1(2\sqrt{t})/\sqrt{t} \\ &= -{}_0\mathrm{F}_1(2; -t) \\ f(-2) &= -t {}_0\mathrm{F}_2(3/2,2; -t^2/4) \\ f(-3) &= \frac{-t^2}{2} {}_0\mathrm{F}_3(4/3, 5/3,2; -t^3/27) \\ f(-n) &= \frac{-t^{n-1}}{(n-1)!} {}_0\mathrm{F}_n \left( \frac{n+1}{n}, \frac{n+2}{n}, \dots, \frac{2n}{n}; -t^n/(n^n) \right) \text{ for } n \in \mathbb{Z}, n < 0 \end{align*} and, for most of these, we've already left the realm of recognizable functions; they don't get more recognizable if the arguments get more interesting. For evaluation by approximation, I suppose it's convenient the definitional series for the ${}_0\mathrm{F}_n$ functions are rapidly converging.
I have an appointment. I'll think about what to do if $\alpha$ is a nonnegative integer.
Note that $\alpha = 0$ gives $\delta(t)/\mathrm{e}$, which has the inconvenience of not actually being a function. (I.e., there's no hope of avoiding a limit, or possibly only defining it weakly, in terms of some integral operator.) And this generically happens for $\alpha \in \mathbb{Z}$, $\alpha \geq 0$. It's not clear what sort of answer you are looking for when the best answer is a distribution.