I need to evaluate the following integral
$$\int_0^{\infty} s^{2-2\alpha}e^{is^2}ds$$ where $\alpha\in[0,\,1]$. I shall call this integral Gamma-function like integrals since it can be brought to Gamma function as follows:
One first replace $s=\sqrt{it}$, then the integral transforms to $$\frac{i^{3/2-\alpha}}{2}\int_0^{-i\infty} t^{1/2-\alpha}e^{-t}dt$$
Finally, one can analytically continue the integral from negative $t$ axis to real $t$ axis to obtain $\frac{i^{3/2-\alpha}}{2}\Gamma(3/2-\alpha)$.
When I throw the two integrals into Mathematica, it does gives me the above result. However, it seems to tells me the integrals only converge on $[1/2,1]$ and diverge on $[0, 1/2]$. Could anybody explain why it is the case?
Update: I think now I understand the convergence by myself. Let us start with the second integral. We denote break the second integral into two parts: $I=I_1+I_2$, where $$I_1=\int_0^{-i\epsilon} t^{\beta}e^{-t}dt$$ $$I_2=\int_{-i\epsilon}^{-i\infty} t^{\beta}e^{-t}dt$$ and $\beta=1/2-\alpha$
To guarantee $\lim_{\epsilon\to0}I_1=0$, we must have $$\beta>-1$$
Now we would like to analytically continue the integral for $I_2$ to the real positive axis. But the condition of doing this is that the integrand should not have any singularity on the fourth quadrant, apart from the small quarter-circle of radius $\epsilon$ near the origin. This analyticity of the integrand is guaranteed only if it vanishes at $t=-i\infty$. So we should require
$$\beta<0$$ which leads to
$$I_2=\int_{\epsilon}^{\infty} t^{\beta}e^{-t}dt$$
When $\beta \in(-1,0)$,i.e., $\alpha\in(1/2,3/2)$, we can express $I$ as $$I=\int_{0}^{\infty} t^{\beta}e^{-t}dt=\Gamma(\beta+1)=\Gamma(3/2-\alpha)$$
The following is partially analytical but relies on Mathematica for the closed-forms. We start by deforming the contour to the ray $\arg s=\phi$ where $0<\phi<\pi/2$. This is accomplished by the substitution $s=re^{i\phi}$, obtaining
\begin{align} I(\phi) &:=\int_0^{\infty e^{i\phi}} s^{2-2\alpha}\exp(i s^2)\,ds\\ &=e^{(1-2\alpha)i\phi}\int_0^\infty r^{2-2\alpha} \exp(i e^{2i\phi}r^2)\,dr\\ &=e^{(1-2\alpha)i\phi}\int_0^\infty r^{2-2\alpha} \exp(i r^2 \cos(2\phi))\exp(-r^2 \sin(2\phi))\,dr.\\ \end{align}
Since $\sin(2\phi)>0$ due to the assumption on $\phi$, the integrand decays like a Gaussian at infinity and so the integral converges. With this in mind, we can ask Mathematica to carry out the integral under this assumption:
Integrate[Exp[I r^2 Exp[2I \[Phi]]]r^(2-2\[Alpha]),{r,0,\[Infinity]},Assumptions->0<\[Phi]<\[Pi]/4]That is, the integral converges so long as the real part of $\alpha$ doesn't exceed $3/2$. Including the prefactor and taking the limit $\phi\to 0^+$, we obtain
FullSimplify[Limit[ConditionalExpression[ 1/2 (-I E^(2 I \[Phi]))^(-(3/2) + \[Alpha])Gamma[3/2 - \[Alpha]], \[Alpha] + Conjugate[\[Alpha]] < 3],\[Phi]->0,Direction->"FromAbove"]]Since Mathematica uses $-i=e^{-i\pi/2}$, we obtain $$I(0^+) =\frac12 e^{i(3/2-\alpha)\pi/2}\tag{Re $\alpha<3/2$}$$ which is seemingly valid on the entirety of $\alpha\in[0,1]$.
A warning, though. On the one hand, for $1/2\leq \alpha<1$ the above formula agrees with the original Mathematica calculation. But the new calculation purports to be valid for $\alpha\leq 1/2$ as well. I'm not sure I understand the issue myself, but note that the exponent $e^{i s^2}$ has modulus 1 along the real line. Hence the rate at which the integrand's modulus grows along the real line is entirely controlled by $s^{2-2\alpha}$: For $\alpha>1/2$ the modulus grows slower than linear, whereas for $\alpha<1/2$ it is faster than linear. Evidently this transition is enough to Mathematica to declare the integral divergent.
By contrast, if we deform the contour just above the real line then the modulus of $e^{i s^2}$ decreases exponentially fast and so there's no issue with convergence. But it does raise the question of what exactly we intend by the original integral: Is it okay to regularize the growth in this way, or is it not? That will probably depend on the original context.