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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 $$\left |\int\limits_1^N e^{ix^\alpha} \ \text{dx} \right| = \mathcal{O}(N^{1-\alpha}).$$

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You've tried expanding the exponential as a series? – J. M. Dec 23 '10 at 15:18
That does not work, I think. – TCL Dec 23 '10 at 15:22
Did you mean $|\int_{1}^{N} e^{ix^{\alpha}}| = \mathcal{O}(N^{\alpha}$)? Otherwise, what definition of BigOh are you using? I presume $i$ is the complex number... Or am I missing something? – Aryabhata Dec 23 '10 at 16:55
@Moron. Yes, that is what I meant. But I thought that is the standard meaning of the O notation. Anyway @Willie gives the correct answer below. – TCL Dec 23 '10 at 17:02
I am not sure it is standard. I will edit the question just in case. – Aryabhata Dec 23 '10 at 20:30
up vote 5 down vote accepted

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.

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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 $|\phi'(x)|\ge\alpha\,N^{\alpha-1}(=\lambda)$.

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Don't quite see that. What are $a,b,\lambda\phi(x)$? – TCL Dec 23 '10 at 18:10
Change of variables. $\int_1^N e^{ix^\alpha} dx = \int_1^2 \frac{N}2 e^{i(N/2)^\alpha y^\alpha} dy$. So take $\phi(y) = y^\alpha$, $\lambda = (N/2)^\alpha$. Use the $k = 1$ version with $\phi'(y) = \alpha y^{\alpha-1}$ monotonically decreasing. Note that the proof of Van der Corput's lemma is also via repeated integration by parts. – Willie Wong Dec 23 '10 at 18:18
I have edited my answer to explain how to apply van der Corput's lemma. – Julián Aguirre Dec 23 '10 at 21:23

$\alpha = 0$ gives the integral as $(N-1)e^{i} = \mathcal{O}(N)$ and $\alpha = 1$ gives the integral to be $\frac{e^{iN}-e^{i}}{i} = \mathcal{O}(1)$.

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That's suggestive but the question didn't include the endpoints in the range of $\alpha$. I also get the integral to be bounded whenever $\alpha<1/3$. – Colin McQuillan Dec 23 '10 at 15:37
@Colin: True. I am hoping that continuity of the integral in $\alpha$ will carry my argument through. – user17762 Dec 23 '10 at 15:39
According to my calculation, it is true for $\alpha=1/2$. – TCL Dec 23 '10 at 15:52

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