Fourier transform of $e^{-a|t|^{\alpha}}$ I was wondering if it is possible to calculate the Fourier transform of $e^{-a|t|^{\alpha}}$ for $a, \alpha > 0$. Thus, can a meaning be given to
$$\int_{-\infty}^\infty e^{i\omega t} e^{-a|t|^{\alpha}}dt $$
in some distribution sense or any other sense
 A: I doubt there is any nice closed-form for generic $\alpha>0$, but that doesn't mean that nothing can be done. By rescaling, we can set $a=1$.

First, your expression gives us good uniform estimate on the size of derivatives of your function. Let
$$\mathcal{A}(\eta)=\int_{\mathbb{R}}e^{-i\eta x}e^{-|x|^\alpha}\,dx.$$
Then, we have that
$$
\begin{aligned}
\left|\mathcal{A}^{(m)}(\eta)\right|&\le \int_{\mathbb{R}} |x|^m e^{-|x|^\alpha}\,dx
\end{aligned}.$$
Restricting the integration range to $[0,\infty)$ followed by the change of variables $x\to x^{1/\alpha}$ and the representation of the Gamma function shows that we have
$$\left|\mathcal{A}^{(m)}(\eta)\right| \lesssim \frac{1}{\alpha}\Gamma\left(\frac{m}{\alpha}+\frac{1}{\alpha}\right)$$
uniformly in $\eta$. Hence the root test shows that $\mathcal{A}$ is real-analytic for $\alpha>1$ on all of $\mathbb{R}$. Of course, with $\alpha=1$, we know that $\mathcal{A}$ is real-analytic on the ball $B(0,1)$, which is consistent with our derivative estimate.

Second, we can good estimates on the growth of $\mathcal{A}$ for large $|\eta|$ assuming $\alpha>1$, which case $\mathcal{A}$ extends to a well-defined function on the entire complex plane. For the general case $\eta=\lambda+i\beta=re^{i\theta}$, $\lambda,\beta\in\mathbb{R}$, you can use the method of steepest descent to understand the behavior $\mathcal{A}$ for $|\eta|\to\infty$. I will focus on the case $\lambda=0, |\beta|\to\infty$ so that we can use the simpler Laplace's method. We can take $\beta>0$ since $\mathcal{A}(i\beta)=\mathcal{A}(-i\beta)$. Hence, we have
$$\mathcal{A}(i\beta)=\int_{\mathbb{R}} e^{\beta x-|x|^\alpha}\,dx$$
We can write
$$\mathcal{A}(i\beta)=\int_0^\infty e^{-\beta x-x^\alpha}\,dx + \int_0^\infty e^{\beta x-x^\alpha}\,dx. $$
The first integrand has no saddle point so we expect it to be exponentially small compared to the second one. Therefore, we have that
$$\mathcal{A}(i\beta)=\big(1+o(1)\big)\int_0^\infty e^{\beta x-x^\alpha}\,dx $$
Now, let's make the change of variables $x\to \beta^{\frac{1}{\alpha-1}}x$ so that we have
$$\mathcal{A}(i\beta)=\beta^{\frac{1}{\alpha-1}}\big(1+o(1)\big)\int_0^\infty \exp\left[-\beta^{\frac{\alpha}{\alpha-1}} (-x+x^\alpha)\right]\,dx.$$
The stationary point of the function $f(x)=-x+x^\alpha$ occurs at $x_\alpha=\alpha^{-\frac{1}{\alpha-1}}$ with $f(x_\alpha)=-\left(1-\frac{1}{\alpha}\right)\alpha^{-\frac{1}{\alpha-1}}$ and $f''(x_\alpha)=\alpha(\alpha-1)\alpha^{-\frac{\alpha-2}{\alpha-1}}$. Therefore, we finally have
$$\mathcal{A}(i\beta)=\beta^{\frac{1}{\alpha-1}} \exp\left(\beta^{\frac{\alpha}{\alpha-1}} |f(x_\alpha)|\right) \sqrt{\frac{2\pi}{|f''(x_\alpha)|}}\big(1+o(1)\big)\qquad \text{ as }\beta\to\infty.$$
