This is from an exercise in Migdal's "Qualitative Methods in Quantum Theory".
For the case for where b>>a, we can arrive at an estimate by re-writing the integral in the following way:
$$ \frac{1}{\sqrt{b}}\int_0^{\infty} e^{-z^2}sin(\frac{a}{b}z^2)dz \approx \frac{a}{b^{3/2}}\int_0^{\infty} e^{-z^2}z^2dz = a\sqrt{\frac{\pi}{16b^3}}. $$
However, I have been stuck trying to show that in the case a>>b, the integral is approximately $\sqrt{\frac{\pi}{8a}}$. I am currently unsure how to treat the sine term, which oscillates quite fast. I considered approximating the integral's value by just considering the first half-cycle of an oscillation, but that didn't seem to bring me any closer to the known estimate.
Any hints/idea what approximations may be used to recover the given estimate when a>>b?
If you use Euler representation of the sine function, the antiderivative is an error function since $$\int e^{-bx^2}\,\sin(ax^2)\,dx=\Im \left( \int e^{ (b-i a)x^2}\,dx\right)$$
As @Travis Willse already wrote in comments $$\int_0^{\infty} e^{-bx^2}\,\sin(ax^2)\,dx=\frac{\sqrt{\pi }}{2 \sqrt[4]{a^2+b^2}}\,\sin \left(\frac{1}{2} \tan ^{-1}\left(\frac{a}{b}\right)\right)$$
Let $a=k b$ and Taylor expand for large values of $k$
$$I=\frac 1 2 \sqrt{\frac{\pi }{2 b k} }\left(1-\frac{1}{2 k}-\frac{3}{8 k^2}+O\left(\frac{1}{k^3}\right)\right)$$ Replace $k$ by $\frac a b$