Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Is there a constant $C$ which is independent of real numbers $a,b,N$, such that

$$\left| {\int_{-N}^N \dfrac{e^{i(ax^2+bx)}-1}{x}dx} \right| \le C?$$

share|cite|improve this question
After integrating by parts and playing around a little bit, you can easily get an estimate like this provided you make an assumption like $|a| > a_0$, $|b| > b_0$. If you want an estimate that makes no such assumption, then I'm not sure there is any reason to expect such an estimate to exist. – Jonathan Aug 3 '12 at 22:33
After some transformation, I need to prove that $$\int_0^N {{e^{ic{x^2}}}\frac{{\sin x}}{x}} dx$$is bounded. But I am stuck there. I am sure that this problem is right since it is an exercise in my book. – Hezudao Aug 4 '12 at 1:05

These integrals are indeed uniformly bounded. As the only effect of changing the signs of $a,b,N$ on the integral is a sign change and/or complex conjugation, we can restrict to $a,b,N\gt0$. Setting $M=Nb$ and $\alpha=a/b^2$, we have $$ \int_{-N}^N\left(e^{i(ax^2+bx)}-1\right)\frac{dx}x=\int_0^M\left(e^{i(\alpha x^2+x)}-e^{i(\alpha x^2-x)}\right)\frac{dx}x. $$ However, the integrand can be written as $2ie^{i\alpha x^2}\frac{\sin x}x$, so is bounded by $2$ in absolute value. Therefore, fixing any constant $K\gt0$, the integral is bounded by $$ \begin{align} 2K+\int_K^{K\vee M}e^{i(\alpha x^2+x)}\frac{dx}x-\int_K^{K\vee M}e^{i(\alpha x^2-x)}\frac{dx}x.&&{\rm(1)} \end{align} $$ I'll prove that this is bounded with the help of a lemma.

Lemma: (van der Corput lemma) If $f\colon[A,B]\to\mathbb{R}$ is convex with $\lvert f^\prime(x)\rvert\ge\lambda\gt0$ and $g\colon[A,B]\to\mathbb{C}$ is differentiable then, $$\left\lvert\int_A^Be^{if(x)}g(x)dx\right\rvert\le\frac2\lambda\left(\lvert g(B)\rvert+\int_A^B\lvert g^\prime(x)\rvert dx\right)$$

As I will only be interested in intervals $[A,B]\subset[K,\infty)$ with $g(x)=1/x$, the bound in the lemma can be written as $$ \begin{align} \left\lvert\int_A^Be^{if(x)}\frac{dx}x\right\rvert\le\frac2{\lambda K}&&{\rm(2)} \end{align} $$ Taking $f(x)=\alpha x^2+x$, this shows that the first integral in (1) is bounded by $2/K$. For the second integral, take $f(x)=\alpha x^2-x$, fix any $0\lt\epsilon\lt1/2$, and first look at value of the integral with integration range restricted to $[0,\epsilon/\alpha]$. In this range, we have $f^\prime(x)\le f^\prime(\epsilon/\alpha)=-(1-2\epsilon)$. So, by inequality (2), this part of the integral is bounded by $2K^{-1}(1-2\epsilon)^{-1}$.

Next, for any fixed $\gamma\gt1/2$, the value of the last integral in (1), restricted to the range $[\epsilon/\alpha,\gamma/\alpha]$ is bounded by $$ \int_{\epsilon/\alpha}^{\gamma/\alpha}\frac{dx}{x}=\log(\gamma/\epsilon). $$ Finally, look at the last integral in (1) restricted to the range $[\gamma/\alpha,\infty)$. As $f^\prime(x)\ge f^\prime(\gamma/\alpha)=(2\gamma-1)$ in this range, inequality (2) shows that this part of the integral is bounded by $2K^{-1}(2\gamma-1)^{-1}$.

Putting these together shows that the set of integrals in the question is bounded above by a constant. The upper bound obtained here is $$ 2K+\frac2K+\frac2{K(1-2\epsilon)}+\frac2{K(2\gamma-1)}+\log(\gamma/\epsilon) $$ for arbitrary positive constants $K,\epsilon,\gamma$ with $\epsilon\lt1/2\lt\gamma$.

share|cite|improve this answer
I put in the optimal constant $c_1=2$ for the van der Corput lemma, which the reference I linked to doesn't mention. That doesn't matter for the existence of an upper bound though. – George Lowther Jul 23 '13 at 1:51

Note that $$ {\int_{-N}^N ({e^{i(ax^2+bx)}-1)\frac{1}{x}dx} }=2i\int_{0}^N e^{iax^2}\frac{\sin bx}{x}dx $$ As I suppose you've already proved.

Maybe, we can approach in the following way

$$I=2i\int_{0}^N e^{iax^2}\frac{\sin bx}{x}dx=-2\int_{0}^{N}\sin ax^2 \frac{\sin bx}{x}dx+2i\int_{0}^{N}\cos ax^2\frac{\sin bx}{x}dx$$ Hence $$\left|I\right|\le 2\sqrt{I_1^2+I_2^2}$$ where $$I_1=\int_{0}^{N}\sin ax^2 \frac{\sin bx}{x}dx,\ I_2=\int_{0}^{N}\cos ax^2 \frac{\sin bx}{x}dx$$

Now, if we can show that both $I_1,\ I_2$ are bounded by some constant independent of $a,b,N$ then $I $ is also bounded.

share|cite|improve this answer
Yes, yes, I was just thinking about that and going to write that in my answer. – Samrat Mukhopadhyay Jul 21 '13 at 5:55
Why is $I_1\le \int_0^N \frac{\sin bx}{x}\,dx$? – 40 votes Jul 21 '13 at 17:29
I wrote that thinking since $\sin ax^2\le 1\ \forall x$. But now, I myself find it dubious because when $\frac{\sin bx}{x}<0$ we can't really apply that to $I_1$. – Samrat Mukhopadhyay Jul 22 '13 at 10:53

we can use theorem for $a \le b$

$ |\int _a^b f(x)dx|\le \int_a^b|f(x)|dx$

$\int _{-N}^N|(e^{i(ax^2+bx)}-1)\frac1x|dx\le \int _{-N}^N|(e^{i(ax^2)}-1)\frac1x|dx\ $ you can play around with the integral to make it more intergreable

share|cite|improve this answer
Unfortunately, the last integral is unbounded when $N\to+\infty$. – Did Feb 16 '13 at 17:28

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.