Asymptotic evaluation of $\int_0^\infty \frac{e^{-\lambda x}}{x} (e^{i\alpha \lambda x^2}-1) dx$ What is the asymptotic expansion of
$$I(\lambda) = \int_0^\infty \frac{e^{-\lambda x}}{x} (e^{i\alpha \lambda x^2}-1) \: dx,$$
as $\lambda \to 0^+$? Here, $\alpha>0$ is a fixed constant. This integral is motivated from physics.
Added: Claude Leibovici provided a nice answer by explicit evaluation of the integral. An argument based on estimates, rather than brute-force calculation, is also welcomed!
What I tried:

*

*The integral is not singular as $x\to 0$, so $I(\lambda)$ is well-defined.

*One may try $e^{i\alpha \lambda x^2}-1 = \sum_{n=1}^\infty \frac{(i\alpha \lambda x^2)^n}{n!}$ and write
$$I(\lambda) = \int_0^\infty \frac{e^{-\lambda x}}{x} \sum_{n=1}^\infty \frac{(i\alpha \lambda x^2)^n}{n!} dx. \tag{1}$$
However, dominated convergence theorem does not seem to apply, since it is hard to find better bounds then
$$|\textrm{(partial sums of $e^{i\alpha \lambda x^2}-1$)}| \leq e^{\lambda x^2}.$$
Furthermore, if one changes the order of limit of (1) by hand, then we obtain a divergent answer:
$$I(\lambda) = \sum_{n=1}^\infty \frac{(i\alpha \lambda)^n}{n!} \int_0^\infty e^{-\lambda x} x^{2n-1} dx = \sum_{n=1}^\infty \frac{(2n-1)!}{n!} \left(\frac{i\alpha}{\lambda}\right)^n .$$

*I also tried the Feynman's trick $\frac{e^{i\alpha \lambda x^2}-1}{x^2} = i\int_0^{\alpha\lambda} e^{ix^2s}ds$ to obtain
$$I(\lambda) = i\int_0^\infty xe^{-\lambda x} \int_0^{\alpha\lambda} e^{ix^2s} ds dx,$$
but I cannot proceed from here. Changing the order of integration seems hard.

 A: After @Gary comment, let
$$x=\frac{t}{\sqrt{\alpha  \lambda }} \qquad \text{and} \qquad \mu=\sqrt{\frac{\lambda }{\alpha }}$$ to make
$$I=\int_0^\infty \frac {e^{-\mu  t}}{t}\left(e^{i t^2}-1\right)\,dt $$
Mathematica returns
$$\color{blue}{I=\log(\mu)+\frac{\pi}{2}   \left(S\left(\frac{\mu }{\sqrt{2 \pi
   }}\right)-C\left(\frac{\mu }{\sqrt{2 \pi }}\right)\right)+}$$ $$\color{red}{\large i}$$
$$\color{blue}{\frac{\pi}{4}   \left(2 \left(C\left(\frac{\mu }{\sqrt{2 \pi
   }}\right)-1\right) C\left(\frac{\mu }{\sqrt{2 \pi }}\right)+2
   \left(S\left(\frac{\mu }{\sqrt{2 \pi }}\right)-1\right)
   S\left(\frac{\mu }{\sqrt{2 \pi }}\right)+1\right)}$$ Expanded as series around $\mu=0$ gives
$$\Re( I)=\log(\mu)-\frac{1}{2} \sqrt{\frac{\pi }{2}} \mu\left(1-\frac{\mu ^2}{12}-\frac{\mu ^4}{160}+O\left(\mu ^6\right) \right)$$
$$\Im (I)=\frac \pi 4--\frac{1}{2} \sqrt{\frac{\pi }{2}} \mu\left(1-\frac{\mu }{\sqrt{2 \pi }}+\frac{\mu ^2}{12}-\frac{\mu
   ^4}{160}+\frac{\mu ^5}{180 \sqrt{2 \pi }}+O\left(\mu ^6\right)\right)$$
