Equality of 2 integrals (complex) let $F(x)=\int_0^x e^{it^k}dt$. Im trying to see why the following holds $$
\lim_{x\to\infty}F(x)=e^{\pi i/2k}\int_0^{\infty}e^{-t^k}dt, $$
and
$$\lim_{x\to-\infty}F(x) = -e^{(-1)^k\pi i/2k}\int_0^{\infty}e^{-t^k}dt$$
Is this application of Cauchy integral formula? Finding suitable contours...
 A: For $r>0$ and $k \in \mathbb{N}, k \ge 2$ consider the homolorphic function
$$
f_k: \Omega_k:=\left\{z \in \mathbb{C}:\ |z|\le r,\ 0\le \arg z\le \theta_k:=\frac{\pi}{2k} \right\} \to \mathbb{C},\ f_k(z)=\exp(iz^k).
$$
Thanks to the Cauchy Integral formula we have
\begin{eqnarray}
0&=&\int_{\partial\Omega_k}f_k(z)\,dz=\underbrace{\int_0^rf_k(x)\,dx}_{A_k(r)}+\underbrace{iR\int_0^{\theta_k}e^{i\theta}f_k(re^{i\theta})\,d\theta}_{B_k(r)}-\underbrace{re^{i\theta_k}\int_0^1f_k((1-t)re^{i\theta_k})\,dt}_{C_k(r)}.
\end{eqnarray}
Since
$$
\sin x\ge \frac{x}{2}\quad \forall x\in [0,\frac{\pi}{2}],
$$
we have
\begin{eqnarray}
|B_k(r)|&\le& r\int_0^{\theta_k}|f_k(Re^{i\theta})|\,d\theta=r\int_0^{\theta_k}\exp(-r^k\sin(k\theta))\,d\theta\\
&\le& r\int_0^{\theta_k}\exp(-kr^k\frac{\theta}{2})\,d\theta=-\frac{2}{kr^{k-1}}\exp(-kr^k\frac{\theta}{2})\Big|_0^{\theta_k}\\
&=&\frac{2(1-e^{-\frac{\pi r^k}{2}})}{kr^{k-1}}.
\end{eqnarray}
It follows that
$$
\lim_{r\to \infty}B_k(r)=0.
$$
Also
\begin{eqnarray}
C_k(r)&=&re^{i\theta_k}\int_0^1f_k((1-t)re^{i\theta_k})\,dt=e^{i\theta_k}\int_0^rf_k(te^{i\theta_k})\,dt=e^{i\theta_k}\int_0^re^{-t^k}\,dt,
\end{eqnarray}
and therefore
$$
\int_0^\infty e^{it^k}\,dt=\int_0^\infty f_k(x)\,dx=e^{i\theta_k}\int_0^\infty e^{-t^k}\,dt=e^{i\frac{\pi}{2k}}\int_0^\infty e^{-t^k}\,dt.
$$
