In this question I am given that the asymptotic expansion of the Airy function for large $z$ is given by
$$Ai(z) = \frac{1}{2}\pi^{-\frac{1}{2}}z^{-1/4}\exp\left(-\frac{2}{3}z^{\frac{3}{2}}\right)\left[1 + O\left(\frac{1}{ z^{3/2}}\right)\right]$$
and the question is to evaluate the first to terms of the integral.
$$I = \int_{0}^x Ai(\xi) \mathrm{d}\xi$$
being told $\int_{0}^\infty Ai(\xi) \mathrm{d}\xi = \frac{1}{3}$. I have attempted this under the assumption that the first two terms are going to be larger than $O\left(\frac{1}{ z^{3/2}}\right)$ (otherwise how can I obtain the first few terms). So for the leading order terms I just need to compute the first couple of terms of
$$\int_{x}^\infty \frac{1}{2}\pi^{-\frac{1}{2}}\xi^{-1/4}\exp\left(-\frac{2}{3}\xi^{\frac{3}{2}}\right) \mathrm{d}\xi. $$
I do this with integration by parts using
$$\int_{x}^\infty \frac{1}{2}\pi^{-\frac{1}{2}}\xi^{-1/4}\frac{\xi^{\frac{1}{2}}}{\xi^{\frac{1}{2}}}\exp\left(-\frac{2}{3}\xi^{\frac{3}{2}}\right) \mathrm{d}\xi.$$
I get $$ I \approx \frac{1}{2}\pi^{-\frac{1}{2}}\exp\left(-\frac{2}{3}x^{\frac{3}{2}}\right)\left[x^{-3/4} + \frac{3}{4}x^{-\frac{9}{4}} \right],$$ which as you can see has the second term too small compared to $\frac{1}{z^{\frac{3}{2}}}$, can anyone see where I am going wrong.
Just to be puristic (not to say pedantic) $$Ai(z) = \frac{1}{2}\pi^{-\frac{1}{2}}z^{-1/4}\exp\left(-\frac{2}{3}z^{\frac{3}{2}}\right)\left[1 + O\left(\frac{1}{ z^{7/4}}\right)\right]$$ For the computation of $$I=\int \frac{1}{2}\pi^{-\frac{1}{2}}z^{-1/4}\exp\left(-\frac{2}{3}z^{\frac{3}{2}}\right)\,dz$$ let $z=\left(\frac{3}{2}\right)^{2/3} t^{4/3}$ to make $$I=\sqrt{\frac{2}{3 \pi }}\int e^{-t^2}\,dt=\frac{\text{erf}(t)}{\sqrt{6}}$$ Back to $z$ and the bounds $$\int_0^x \frac{1}{2}\pi^{-\frac{1}{2}}z^{-1/4}\exp\left(-\frac{2}{3}z^{\frac{3}{2}}\right)\,dz=\frac{1}{\sqrt{6}}\text{erf}\left(\sqrt{\frac{2}{3}} x^{3/4}\right)$$ $\frac{1}{\sqrt{6}} \sim 0.41$ which is $20$% higher than $\frac 13$.