= 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> gives
$$
\boxed{
\lim_{R\to\infty}\int_{-R}^R e^{-itu^2}du
=e^{-i\pi/4}\sqrt{\frac\pi t}.}
$$
Similarly, 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
$$
\boxed{
\lim_{R\to\infty}\int_0^R e^{-itu^3}du
=e^{-i\pi/6}t^{-1/3}\Gamma\left(\frac43\right).}
$$
For fixed $\alpha$, consider the complex phase
$$
\Phi(\theta)=x\sin\theta-\alpha\theta.
$$
Its unique interior stationary point is
$$
\theta_0=\arccos(\alpha/x)
=\frac\pi2-\frac\alpha x+O(x^{-3}),
$$
where
$$
\Phi(\theta_0)=x-\frac{\pi\alpha}{2}+O(x^{-1}),
\qquad
\Phi''(\theta_0)=-x+O(x^{-1}).
$$
The 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>:
$$
\boxed{
Q_\alpha(x)\sim
\sqrt{\frac2{\pi x}}
\cos\left(x-\frac{\pi\alpha}{2}-\frac\pi4\right).}
$$
For $Q_x(x)$ the phase is $x(\sin\theta-\theta)$. Its stationary point has merged with the endpoint $\theta=0$, where
$$
\sin\theta-\theta=-\frac{\theta^3}{6}+O(\theta^5).
$$
Use the <cubic endpoint stationary-phase scaling> $\theta=(6/x)^{1/3}u$. The other endpoint and the region away from zero contribute only $O(x^{-1})$, so
$$
\begin{aligned}
Q_x(x)
&\sim\frac1\pi\left(\frac6x\right)^{1/3}
\int_0^\infty\cos(u^3)du\\
&=\frac{6^{1/3}}{\pi x^{1/3}}
\Gamma\left(\frac43\right)\cos\frac\pi6.
\end{aligned}
$$
Therefore
$$
\boxed{
Q_x(x)\sim
\frac{\sqrt3 6^{1/3}}{2\pi}
\Gamma\left(\frac43\right)x^{-1/3}.}
$$
Back to article page
