MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

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

Are there any known necessary or sufficient conditions such that $$\lim_{a\rightarrow \infty}\int_{-1}^1f(t)^a\cos(at)dt=0$$ where $f:[-1,1]\rightarrow[1,\infty)$ is an even smooth concave real function such that $f(-1)=f(1)=1$?

share|cite|improve this question

Sorry for the late reply (for the last month I hardly had any time for anything like MO). The answer (somewhat vague) is that the curve $t\mapsto (t,\log f(t))$ ($-1\le t\le 1$) has to be the image of the upper half circle under some analytic in $\mathbb C\setminus[(-\infty,-1]\cup[1,+\infty)]$ mapping $F$, which is symmetric ($F(\bar z)=\overline{F(z)}$), one to one in the unit disk, and whose derivative $F\\,'$ has decent boundary behavior. The "decent boundary behavior" is the vague part here. Unfortunately, I cannot make it less vague unless someone tells me how exactly to recognize the distributions on $[-1,1]$ whose Fourier transform tends to $0$. It is clear that they are necessarily fairly tame (of not more than the first order, etc.) and that all $L^1$-functions are there but where exactly you are in between is a mystery to me. On the crudest level, it tells you that $f$ must be analytic.

Let me know if such description is of any interest to you. If it is, I'll post the details.

EDIT: OK, here go the details. It is a somewhat long story, so I may need more than one patch of free time to type it. I apologize in advance for bumping this thread. Also, since the integral against $\sin at$ is zero, we can just as well talk about the full Fourier transform, i.e., the integration against $e^{-iat}$

1) It is actually quite surprising that such functions exist at all. After all, the jump discontinuity normally means that the best rate of decay of the Fourier transform is $1/a$ and that slow rate of decay is played against the exponential growth of the integrand. So, I'll start with constructing one such function. It'll be easier to work with $g(t)=\log f(t)$, which is a smooth non-negative function with endpoint values $0$. Put $g(t)=\delta(1-t^2)$ with small $\delta>0$. Then the integral can be written as the path integral $\int_\gamma \frac{dt}{dz}e^{-iaz}\\,dz$ where $\gamma$ is the curve $t\mapsto z(t)=t+ig(t)$. Note that $z(t)=t+i\delta(1-t^2)$ is an analytic function of $t$ and for small $\delta>0$, it is invertible in a fairly large disk. Thus, we can talk about its analytic branch $t(z)$ that coincides with $\Re z$ on $\gamma$ and is analytic in (a neighborhood of) the region $D$ bounded by $[-1,1]$ and $\gamma$. So, $t'(z)$ is also analytic there and we can shift the contour of integration from $\gamma$ to $[-1,1]$, which results in the representation $\int_{-1}^1 t'(z)e^{-iaz}\\,dz$, which is just the ordinary Fourier transform of the integrable (and even smooth) function $t'(z)$ restricted to $[-1,1]$, so the integral, indeed, tends to $0$ in this case.

2) What I'd like to show now is that this contour integral representation and the possibility to shift the contour is the only possible reason for this effect. The starting point is that if the integral is bounded on the entire real line (the boundedness on the negative semi-axis is trivial and the boundedness on the positive one is less than what has been requested), then there exists a distribution $T$ supported on $[-1,1]$ such that the integral equals $\langle T, e^{-iat}\rangle$ for all $a\in\mathbb C$ (that is just a version of Paley-Wiener). Thus, the difference $\langle T, e^{-iat}\rangle-\int_\gamma \frac{dt}{dz}e^{-iaz}\\,dz$ vanishes for all $a\in \mathbb C$. Now, the linear span of functions $e^{-iaz}$ is dense in the space of functions analytic in any fixed neighborhood of $D$ meaning that $\langle T, \psi(t)\rangle-\int_\gamma \frac{dt}{dz}\psi(z)\\,dz=0$ for every function analytic in some neighborhood of $D$.

We will take $\psi(z)=\frac{1}{z-\zeta}$ with $\zeta\notin D$ and get the Cauchy integral plus something analytic in $\mathbb C\setminus[-1,1]$ vanish outside $D$. Note that this Cauchy integral and the distribution part are also well-defined for $\zeta\in D$ and give an analytic function of $\zeta$ there. Moreover, by Plemelj's jump formulae, the boundary values of that function on $\gamma$ are just $\frac{dt}{dz}$ (up to $2\pi i$ and $\pm$, which we aren't concerned with here). The upshot is that $\frac{dt}{dz}$ has an analytic extension to $D$ continuous up to $\gamma$ (here we use that the curve is assumed to be of some decent smoothness; otherwise we'll have to sing a long song of non-tangential boundary values a.e., etc.)

The behavior on $[-1,1]$ may be more complicated in general and the boundary values there exist only in the sense of distributions. The possibility of analytic continuation to the open domain $D$ guarantees only the possibility to shift the contour to something hovering as low over $[-1,1]$ as we wish, i.e., to the subexponential growth of the integral (for which it is necessary and sufficient). However, if you settle for some more reasonable class than $C_0$, say, $L^2$, then $T$ will be just an $L^2$ function and you'll have the classical theory of boundary values that will allow you to show that our distribution is, indeed, the boundary value of the analytic extension of $\frac{dt}{dz}$ and the reason for smallness of the integral is the possibility of the ordinary contour shift. I have no idea what you are going to use all this for, so I prefer to avoid the discussion of all those technical issues. Instead, I'll discuss in detail what the possibility of this analytic extension of the derivative means for the curve $\gamma$ itself.

3) Let $Q$ be the lower unit half-disk $\{z:|z|<1,\Im z<0\}$. Let $\varphi$ be the conformal mapping from $Q$ to $D$ such that the interval $[-1,1]\subset\partial Q$ is mapped to $\gamma$ and the lower semicircle is mapped to $[-1,1]$. The derivative $\varphi'$ is a continuous up to the boundary (except for the points $-1,1$ where it has an easy to control power singularity) non-vanishing function in $D$ (here we use reasonable smoothness of $f$ again). Note that after the composition with $\varphi$, the function $\frac{dt}{dz}$ on $\gamma$ becomes $\frac{(\Re\varphi)'}{\varphi'}$ on $[-1,1]$. This should be extendable analytically to the lower half-disk with "decent boundary values". Since $\varphi'$ has such extension, we conclude that so does $(\Re\varphi)'$. But this function is real-valued, so the Schwarz reflection principle applies and we conclude that it extends analytically to the entire unit disk. Let $(\Re\varphi)'=F$ where $F$ is a symmetric analytic function in the unit disk. The function $\varphi'-F$ is purely imaginary on $[-1,1]$ and extends analytically to the lower half-circle. Thus, using the reflection principle again, we conclude that $\varphi'=F+iG$ where $F,G$ are symmetric analytic functions in the unit disk and $F$ has decent boundary values on the lower semicircle. Thus, $\varphi'$ and $\varphi$ are analytic in the unit disk. Moreover, since $F+iG=\varphi'$ is nice on the lower semicircle, $G$ also has decent boundary values there and therefore, after reflecting, we see that $\varphi'$ has decent boundary values on the upper semicircle. To get the proclaimed description, it suffices now to map the unit disk to the upper half-plane so that the lower semicircle is mapped to $[-1,1]$ and use the reflection principle again for the last time.

That's it (modulo minor technicalities that I swept under the rug, but, as I said, to get into those would make no sense without knowing what exactly you are after).

share|cite|improve this answer
Thank you for your answer. More details would definitively be helpful. – Roland Bacher Mar 9 '12 at 16:38

Here is a possible beginning. Note first that

$$ \int_{-1}^1 f(t)^a \cos(at) dt= 2I_a:= 2\int_{0}^1 f(t)^a \cos(at) dt. $$

Hence, it suffices to investigate $I_a$. Let me first assume that $f'(t) <0$ on $(0,1)$. (Note that if $f'(t_0)=0$ for some $t_0\in (0,1)$ then $f'(t)=0$ on $[0,t_0]$.) This means that the map $t\mapsto f$ is one-to-one. We regard $t$ as a function of $f$. Then the change in variables formula implies.

$$I_a= \int_1^{f(0)} f^a \cos(a t)\frac{dt}{df} df $$

I can make this formula friendlier to the 21st century mathematician by changing notations,

$$t \longleftrightarrow \phi,\;\;\; f \longleftrightarrow x$$

and we can rewrite the above as

$$ I_a= \int_1^{x_0} x^a\frac{d\phi}{dx} \cos( a \phi(x) ) dx = \frac{1}{a}\int_1^{x_0} x^a \frac{d}{dx}\Bigl( \sin\bigl(\; a\phi(x)\;\bigr) \Bigr)dx$$

$$=\frac{1}{a}\Bigl( x^a\sin\bigl( a\phi(x)\bigr)\;\Bigr)\Bigr|^{x_0}_1- \int_1^{x_0}x^{a-1}\sin\bigl(\; a\phi(x) \;\bigr)dx. $$

Now observe that $\phi(x_0) =0$, $\phi(1)=1$, so the first term above goes to zero as $a\to\infty$.

At this point it may be useful to look in some books on asymptotics of integrals. A good place to start is

Bleistein & Handelsman: Asymptotic expansions of Integrals, Dover

Also you need to keep in mind that

$$\frac{d\phi}{dx}< 0,\;\;\forall x\in (1,x_0)$$

$$\lim_{x\nearrow x_0} \frac{d\phi}{dx}=-\infty. $$

share|cite|improve this answer
Thank you for the reference, I will check if it is useful. – Roland Bacher Feb 16 '12 at 17:33

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.