 Solution

Rotate the contour for the Gaussian integral through angle $-\pi /4$; a small positive damping factor justifies the deformation and can then be removed. After scaling by $t^{-1/2}$, the oscillatory Gaussian integral with negative phase givesSimilarly, first scale $u=t^{-1/3}v$ and then rotate the positive ray through angle $-\pi /6$. Since ${\int }_0^{\infty }{e}^{-v^3}dv=\Gamma (4/3)$, the cubic oscillatory Gamma integral gives

For fixed $\alpha$, consider the complex phaseIts unique interior stationary point iswhereThe endpoint contributions are $O(x^{-1})$, while the one-dimensional stationary-phase formula has order $x^{-1/2}$. Taking its real part proves the stationary-phase asymptotic for the generalized Bessel cosine integral:

For $Q_x(x)$ the phase is $x(\sin \theta -\theta )$. Its stationary point has merged with the endpoint $\theta =0$, whereUse the cubic endpoint stationary-phase scaling $\theta =(6/x{)}^{1/3}u$. The other endpoint and the region away from zero contribute only $O(x^{-1})$, soTherefore