Region of convergence of $\int_0^{\infty} x^s e^{-\frac{|\log(x)|^k}{2}}dx$ where $s \in \mathbb{C}$. I am interested in finding the region of convergence of the integral
\begin{align}
\int_0^{\infty} x^s e^{-\frac{|\log(x)|^k}{2}}dx,
\end{align}
where $s \in \mathbb{C}$ and $k>0$.   I would like to solove this for all $k$ but I am espeicially interesed in $0<k<1$.
How do we approach this type of problem? 
Note for $k=1$ this integral has been solved here. 
However, that approach does not generalize.
 A: Enforcing the substitution $x=e^u$ yields
$$\begin{align}
\int_0^\infty x^s e^{-\frac12 |\log(x)|^k}\,dx&=\int_{-\infty}^\infty e^{(s+1)u-\frac12 |u|^k}\,du\\\\
&=\int_{-\infty}^0 e^{(s+1)u-\frac12 (-u)^k}\,du+\int_{0}^\infty e^{(s+1)u-\frac12 u^k}\,du\\\\
&=\int_0^\infty e^{-\frac12 u^k-(s+1)u}\,du+\int_{0}^\infty e^{(s+1)u-\frac12 u^k}\,du\tag 1
\end{align}$$

For each fixed $k\in (0,1)$, the first integral on the right-hand side of $(1)$ converges when $\text{Re}(s)\ge -1$.  To see this, we note that when $\text{Re}(s)\ge -1$, $\left(\text{Re}(s)+1\right)u+\frac12u^k\ge \frac12 u^k$ and $\int_0^\infty e^{-\frac12 u^k}\,du$ converges.

For each fixed $k\in (0,1)$, the first integral on the right-hand side of $(1)$ diverges when $\text{Re}(s)< -1$.  To see this, we note that when $\text{Re}(s)< -1$, $\left(\text{Re}(s)+1\right)u+\frac12u^k< -\frac12 u^k$ for $u>u_0=\left(\frac{1}{|\text{Re}(s)+1|}\right)^{1/(1-k)}$.  And the integral $\int_{2u_0}^\infty e^{\frac12 u^k}\,du$ diverges.

For each fixed $k\in (0,1)$, the second integral on the right-hand side of $(1)$ converges when $\text{Re}(s)\le -1$.  To see this, we note that when $\text{Re}(s)\le -1$, $\left(\text{Re}(s)+1\right)u-\frac12u^k\le -\frac12 u^k$ and $\int_0^\infty e^{-\frac12 u^k}\,du$ converges.

For each fixed $k\in (0,1)$, the second integral on the right-hand side of $(1)$ diverges when $\text{Re}(s)>-1$.  To see this, we note that when $\text{Re}(s)>-1$, $(\text{Re}(s)+1)u-\frac12 u^k>\frac12 u^k$ for $u>u_1=\left(\frac{1}{\text{Re}(s)+1}\right)^{1/(1-k)}$.  And the integral $\int_{2u_1}^\infty e^{\frac12 u^k}\,du$ diverges.


Putting everything together, for each $k\in (0,1)$, the integral of interest converges when $\text{Re}(s)=-1$ and diverges otherwise.



Closed Form Solution for $\displaystyle s=-1$

When $s=-1$, we see that 
$$\begin{align}
\int_0^\infty x^{-1}e^{-\frac12 |\log(x)|^k}\,dx&=2\int_0^\infty e^{-\frac12 u^k}\,dx
\end{align}$$
Enforcing the substitution $u=x^{1/k}$ reveals
$$\begin{align}
\int_0^\infty x^{-1}e^{-\frac12 |\log(x)|^k}\,dx&=\frac{2}{k}\int_0^\infty e^{-\frac12 x}\,x^{1/k-1}\,dx\\\\
&=\frac{2^{1/k+1}}{k}\int_0^\infty e^{-x}x^{1/k-1}\,dx\\\\
&=\frac{2^{1/k+1}}{k}\,\Gamma(1/k)
\end{align}$$
