Proving that $\int_{0}^{+\infty}e^{ix^n}\text{d}x=\Gamma\left(1+\frac{1}{n}\right)e^{i\pi/2n}$ I've shown that for all $p \in \mathbb{R}^{*+}$
$$\int_{0}^{+\infty}e^{-x^p}\text{d}x=\Gamma\left(1+\frac{1}{p}\right)$$
And I want to show that 
$$
\int_{0}^{+\infty}e^{ix^n}\text{d}x=\Gamma\left(1+\frac{1}{n}\right)e^{i\pi/2n}
$$
Is that possible with a simple change of variable ? 
 A: $x=u^{1/n}$, $dx=\frac1n u^{1/n-1}\,du$
$$\begin{align}
\int_0^{\infty} e^{ix^n}\,dx&\overbrace{=}^{x\mapsto u^{1/n}}\frac1n \int_0^\infty e^{iu}u^{1/n-1}\,du\\\\
&\overbrace{=}^{u\mapsto ix}\frac1n \int_0^{i\infty} e^{-x} (ix)^{1/n-1}\,i\,dx\\\\
&=\frac1n e^{i\pi/(2n)}\int_0^{i\infty} x^{1/n-1}e^{-x}\,dx\tag1\\\\
&=\frac1n e^{i\pi/(2n)}\int_0^{\infty} x^{1/n-1}e^{-x}\,dx\tag2\\\\
&=e^{i\pi/(2n)}\frac1n \Gamma\left(\frac1n\right)\tag3\\\\
&=e^{i\pi/(2n)}\Gamma\left(1+\frac1n\right)\tag 4
\end{align}$$
where in going from $(1)$ to $(2)$, we used Cauchy's Integral Theorem to deform the contour back to the real line, and in going from $(3)$ to $(4)$ we used the function equation $\Gamma(1+x)=x\Gamma(x)$.

Note in using Cauchy's Integral Theorem to go from $(1)$ to $(2)$, we need to account for a branch point at $z=0$ in the complex plane.  We chose to cut the plane along the negative real axis.  
Then, we can write
$$\begin{align}
0&=\oint_C e^{-z} z^{1-1/n}\,dz\\\\
&=\int_{i\epsilon}^{iR} x^{1/n-1}e^{-x}\,dx-\int_{\epsilon}^R x^{1/n-1}e^{-x}\,dx\\\\
&+\int_0^{\pi/2}(\epsilon e^{i\phi})^{1/n-1}e^{-\epsilon e^{i\phi}}\,i\epsilon e^{i\phi}\,d\phi-\int_0^{\pi/2}(R e^{i\phi})^{1/n-1}e^{-R e^{i\phi}}\,iR e^{i\phi}\,d\phi\tag5
\end{align}$$ 
where we exploited the fact that $z^{1/n-1}e^{-z}$ is analytic in and on $C$ (i.e., We have excluded the branch point and corresponding cut from $C$).
It is straightforward to show that the third and fourth integrals on the right-hand side of $(5)$ vanish as $\epsilon\to 0$ and $R\to \infty$.  
Hence , we deduce the coveted relationship
$$\int_{0}^{I\infty} x^{1/n-1}e^{-x}\,dx=\int_{0}^\infty x^{1/n-1}e^{-x}\,dx$$
as was to be shown.
