How do you integrate $\int_0^\infty \exp(it^k)\,\mathrm dt$ for $k \in \Bbb N$?

Make the angle of your pizza slice equal to $\pi/(2 k)$. Then we have

$$\oint_C dz \, e^{i z^k} = \int_0^R dx \, e^{i x^k} + i R \int_0^{\pi/(2 k)} d\phi \, e^{i \phi} \, e^{-R^k \sin{k \phi}} e^{i R^k \cos{k \phi}} + e^{i \pi/(2 k)} \int_{\infty}^0 dt \, e^{-t^k} $$

which is zero due to Cauchy's theorem. Use the fact that $\sin{k \phi} \ge (2/\pi) k \phi$ over $\phi \in [0,\pi/(2 k)]$, so that

$$\left | i R \int_0^{\pi/(2 k)} d\phi \, e^{i \phi} \, e^{-R^k \sin{k \phi}} e^{i R^k \cos{k \phi}}\right| \le R \int_0^{\pi/(2 k)} d\phi \, e^{-(2 k R^k/\pi) \phi}\le \frac{\pi}{2 k R^{k-1}} $$

which vanishes as $R \to \infty$. In this limit, then,

$$\int_0^{\infty} dx \, e^{i x^k} = e^{i \pi/(2 k)} \int_0^{\infty} dx \, e^{-x^k} = e^{i \pi/(2 k)} \Gamma{\left ( 1+ \frac{1}{k}\right)} $$