Let $x>1$ and $f(x)=\int_{0}^{+\infty}e^{it^{x}}dt$. Does this integral have a closed form ?
Fist point, the integral converges. Indeed let $u=e^{it^{x}}$ and $v=\frac{-i}{x}t^{1-x}$ we have by integration by parts $$\lim_{T\rightarrow+\infty}\int_0^T=\int_0^1 e^{it^x}dt+\frac{i}{x}e^i+\frac{1-x}{x}\int_1^{+\infty}\frac{e^{it^x}}{t^x}dt$$ One can also prove that $f(x)\sim_{+\infty}1.$
For integral $$f(x)=\int_{0}^{+\infty}e^{it^{x}}dt$$
We first change variable from $t$ to $v=-i t^x$ so that $t=v^{1/x}i^{1/x}$ and $dt=i^{1/x}(1/x)v^{-1+1/x}dv$.
The integral now becomes
$$f(x)=i^{1/x}(1/x)\int_0^{+\infty}e^{-v}v^{-1+1/x}dv......(1)$$
Mathematica 7.0 gives:
$$f(x)=(1/x)\exp\left(\frac{i\pi}{2x}\right)\Gamma(1/x)$$
The definition of the $\Gamma(z)$ function is given by:
$$\Gamma(t)=\int_0^{+\infty}e^{-v}v^{t-1}dv......(2)$$
Comparison of (1) against (2) leads to the same result:
$$f(x)=(1/x)i^{1/x}\Gamma(1/x)=(1/x)\exp\left(\frac{i\pi}{2x}\right)\Gamma(1/x)......(3)$$