= Solution
Near zero, write the amplitude as $t^{-1/4}(1+t)^{-1/4}=\sum_{m\geq0}\binom{-1/4}{m}t^{m-1/4}$. The large parameter localizes the integral at $t=0$. Termwise integration of the local expansion gives
$$
\int_0^\infty e^{-kt^2}t^{m-1/4}dt
=\frac12 k^{-3/8-m/2}\Gamma\left(\frac38+\frac m2\right).
$$
Consequently
$$
\boxed{\alpha=\frac38,\quad\beta=\frac12,\quad
a_m=\frac12\binom{-1/4}{m}\Gamma\left(\frac38+\frac m2\right)
=\frac{(-1)^m(1/4)_m}{2m!}\Gamma\left(\frac38+\frac m2\right)}.
$$
Here $(1/4)_m$ is the <rising factorial>. This is an asymptotic series, not an assertion that the full integrated Taylor series converges.
The needed result is <Watson's lemma>: if a locally integrable amplitude has an expansion $g(s)\sim\sum c_ms^{\rho_m-1}$ as $s\downarrow0$, with $\rho_m>0$ increasing and suitable growth at infinity so its Laplace integral converges, then $\int_0^\infty e^{-ks}g(s)ds\sim\sum c_m\Gamma(\rho_m)k^{-\rho_m}$, with the corresponding finite-order remainder bounds. Apply it after $s=t^2$, when $g(s)=\tfrac12s^{-5/8}(1+\sqrt s)^{-1/4}$ and $\rho_m=3/8+m/2$. The tail is exponentially small after any fixed positive cutoff.
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