I have the following integral: $$f(x) = \int_{1}^{x} \dfrac{t^{-\alpha}}{(x - t)^{\frac{3}{2}} } \exp{\left( \dfrac{-A^2 {t}^2}{(x - t)} \right)} dt, A > 0, \alpha > 0$$
I want to calculate the log-log slope of $f(x)$ for $x$ large, say $10^4$. I can see numerically (after numerical integration and plotting) that for certain range of values of $A$ and $\alpha$, the log-log slope (abs value) of $f(x)$ seems to scale as $1 + \alpha/2$, other times like $1 + \alpha$.
I would like to see if asymptotics can get me anywhere towards that, but this integral seems to not fit any of the standard forms like Laplace. Any ideas ?
For fixed $A$ and $\alpha$, the asymptotic can be found in a way similar to this; after the change of variables $\xi = t/(x - t)$, the leading term is determined by the maximum of the exponent at $\xi = 0$ but the integration range is $[1/(x-1), \infty)$: $$f(x) = \int_{1/(x - 1)}^\infty \frac 1 {\sqrt {x (\xi + 1)}} \left( \frac {x \xi} {\xi + 1} \right)^{-\alpha} e^{-A^2 x \xi^2 / (\xi + 1)} d\xi \sim \\ x^{-\alpha - 1/2} \int_{1/(x - 1)}^\infty \xi^{-\alpha} e^{-A^2 x \xi^2} d\xi = \\ \frac 1 2 A^{\alpha - 1} x^{-\alpha/2 - 1} \Gamma \!\left( \frac {1 - \alpha} 2, \frac {A^2 x} {(x - 1)^2} \right) \sim \\ \frac 1 2 A^{\alpha - 1} x^{-\alpha/2 - 1} \Gamma \!\left( \frac {1 - \alpha} 2, \frac {A^2} x \right) \sim \\ \cases { \frac 1 2 A^{\alpha - 1} \Gamma \!\left( \frac {1 - \alpha} 2 \right) x^{-\alpha/2 - 1} & $\alpha < 1$ \\ \frac 1 2 x^{-3/2} \ln x & $\alpha = 1$ \\ \frac 1 {\alpha - 1} x^{-3/2} & $\alpha > 1$ \\}, \quad x \to \infty.$$