Asymptotic behavior of $\int_0^\infty e^{i\alpha \xi^2} e^{-\beta \xi} e^{ix \xi} d\xi$

Question

For $\alpha\in\mathbb R$ and $\beta>0$, consider the function $$f(x) = \int_0^\infty e^{i\alpha \xi^2} e^{-\beta \xi} e^{ix \xi} \: d\xi,$$ where $x\in\mathbb R$. In particular, what is the asymptotic behavior of $f(x)$ as $x\to\pm\infty$?

Mathematica says that the above integral is

DawsonF[(-I x + \[Beta])/(2 Sqrt[I \[Alpha]])]/Sqrt[I \[Alpha]]

which asymptotically grows (exponentially!) for $x \to \textrm{sgn}(\alpha) \,\infty$. However, I do not trust this result, since $f$ has finite $L^2$-norm $$\| f \|_2^2 = 2\pi \int_0^\infty e^{-2\beta \xi} \: d\xi = \frac\pi\beta$$ by the Plancherel's theorem.

Answer 1

Completing the square $$\alpha \xi ^2+ (x+i \beta )\,\xi=\left(\sqrt{\alpha }\, \xi +\frac{x+i \beta }{2 \sqrt{\alpha }}\right)^2-\frac{(x+i \beta )^2}{4 \alpha }$$

Let $$\xi=\frac{t}{\sqrt{\alpha }}-\frac{x+i \beta }{2 \alpha }$$ and the antiderivative is $$I=\frac{1}{\sqrt{\alpha }}\exp\left( -\frac{i (x+i \beta )^2}{4 \alpha } \right)\int e^{i t^2}\,dt$$ $$\int e^{i t^2}\,dt=-\frac{1}{2} (-1)^{3/4} \sqrt{\pi }\, \text{erfi}\left((-1)^{1/4} t\right)$$

Back to $x$ and the bounds, the definite integral is

$$I=\frac{(-1)^{1/4} \sqrt{\pi }}{2 \sqrt{\alpha }}\exp\left( -\frac{i (x+i \beta )^2}{4 \alpha } \right) \left(1+\text{erf}\left(\frac{(-1)^{3/4} (x+i \beta )}{2 \sqrt{\alpha }}\right)\right)$$ whose asymptotics is $$I\sim \frac{i}{x}+\frac{\beta }{x^2}+\frac{2 \alpha -i \beta ^2}{x^3}-\frac{i \beta \left(6 \alpha -i \beta ^2\right)}{x^4}+O\left(\frac{1}{x^5}\right)$$

Answer 2

Choose $\alpha=0$ and $\beta=1$. So we want to find the asymptotic behavior of $$\int_0^\infty e^{-\xi}e^{ix\xi}d\xi$$This is really the laplace transform of $e^{ix\xi}$. Writing this imaginary exponential with its real and imaginary parts, we get that the integral is equal to $$\frac{1}{1+x^2}+i\frac{x}{1+x^2}$$So as $x$ approaches infinity, the integral approaches $0$. The process should be similar for any $\alpha$ or $\beta$ with the limitations given in the OP.