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

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

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  • $\begingroup$ 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. $\endgroup$
    – Jonathan
    Commented Aug 3, 2012 at 22:33
  • $\begingroup$ 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. $\endgroup$
    – Summer
    Commented Aug 4, 2012 at 1:05

3 Answers 3

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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 $\epsilon\lt1/2\lt\gamma$.

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  • $\begingroup$ 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. $\endgroup$ Commented Jul 23, 2013 at 1:51
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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.

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  • $\begingroup$ Yes, yes, I was just thinking about that and going to write that in my answer. $\endgroup$ Commented Jul 21, 2013 at 5:55
  • $\begingroup$ Why is $I_1\le \int_0^N \frac{\sin bx}{x}\,dx$? $\endgroup$
    – 40 votes
    Commented Jul 21, 2013 at 17:29
  • $\begingroup$ 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 $I_1$. $\endgroup$ Commented Jul 22, 2013 at 10:53
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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

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    $\begingroup$ Unfortunately, the last integral is unbounded when $N\to+\infty$. $\endgroup$
    – Did
    Commented Feb 16, 2013 at 17:28

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