Is this integral of order $N^{\alpha}$? Is  $$\left |\int\limits_1^N e^{ix^\alpha} \ \text{dx} \right| = \mathcal{O}(N^\alpha)?$$ 
where $\displaystyle 0\lt \alpha \lt 1$ and $\displaystyle i=\sqrt{-1}$.
${\bf Edit.}$ As Willie Wong shows below, the correct equation is 
$$\left |\int\limits_1^N e^{ix^\alpha} \ \text{dx} \right| = \mathcal{O}(N^{1-\alpha}).$$
 A: I think what you should get is, for each fixed $\alpha$, asymptotically $O(N^{1-\alpha})$. 
Here's why: do the change of variables $y = x^\alpha$. Your integral becomes
$$ \alpha \int_1^{N^\alpha} y^{\frac{1-\alpha}\alpha} e^{iy} dy $$
Now integrate by parts to lower the exponent on $y$. The first step would be
$$ = \frac{\alpha}{i}\left( N^{1-\alpha} e^{i N^\alpha} - e^i \right) - \frac{1-\alpha}{i} \int_1^{N^\alpha} y^{\frac{1-2\alpha}\alpha} e^{iy} dy$$
Repeating the process $M = \lceil \frac{1}{\alpha} \rceil$ number of times, you get a string of terms that estimates to 
$$ = O(N^{1-\alpha} + N^{1-2\alpha} + \ldots + N^{1-(M-1)\alpha}) + O(\int_1^{N^\alpha} y^{\frac{1-M\alpha}{\alpha}} e^{iy} dy) $$
Since $M\alpha \geq 1$, the final integral is dominated by a constant. Noting that for large $N$, the first term inside the first $O(\cdot)$ dominates, you get the result. 
A: Van der Corput's lemma gives
$$
\left|\int_1^N e^{ix^\alpha}dx\right|\le \frac{C}{\alpha}\,N^{1-\alpha},
$$
where the constant $C$ is independent of $\alpha$ (and of $N$).
Edited to answer TCL's comment
Van der Corput's lemma
If $\phi\colon(a,b)\to\mathbb{R}$ is smooth and $|\phi^{(k)}(x)|\ge1$ for $a<x<b$, then
$$
\left|\int_1^N e^{i\lambda\phi(x)}dx\right|\le C_k\lambda^{-1/k}.
$$
This valid if $k\ge2$ or if $k\ge1$ and $\phi'$ is monotone.
It is clearly equivalent to
If $\phi\colon(a,b)\to\mathbb{R}$ is smooth and $|\phi^{(k)}(x)|\ge\lambda$ for $a<x<b$, then
$$
\left|\int_a^b e^{i\phi(x)}dx\right|\le C_k\lambda^{-1/k}.
$$
To aply it to $\int_1^N e^{ix^\alpha}dx$, choose $a=1$, $b=N$, $\phi(x)=x^\alpha$, and $k=1$. Then $\phi'(x)=\alpha\,x^{\alpha-1}$ is decrasing, so that $|\phi'(x)|\ge\alpha\,N^{\alpha-1}(=\lambda)$.
A: $\alpha = 0$ gives the integral as $(N-1)e^{i} = \mathcal{O}(N)$ and $\alpha = 1$ gives the integral to be $\frac{e^{iN}-e^{i}}{i} = \mathcal{O}(1)$.
