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Can one describe all the real polynomials $P(x)$ such that the following integrals converge: $$ \int_0^{\infty} \sin(P(x))dx, \int_0^{\infty} \cos(P(x))dx ? $$

Among special cases are such celebrities as Airy and Fresnel functions.

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    $\begingroup$ I think that this would get a better response if it would come as a question and not a command and also accompanied by some motivation and more background. $\endgroup$
    – Dirk
    Oct 21, 2014 at 6:53
  • $\begingroup$ Dirk, is not it a question-- ask to find all such polynomes? And to generalize conditions for famous Airy and Fresnel functions--is not a motivation? $\endgroup$
    – Sergei
    Oct 21, 2014 at 7:02
  • $\begingroup$ I think some indication of what your thoughts are might help also. $\endgroup$
    – Igor Rivin
    Oct 21, 2014 at 8:57
  • $\begingroup$ Now I notice the question mark behind the formula. Maybe it's just a nitpick, but "Specify all…?" does not sound like a question to me (in contrast to "What are all…?"). Generalization per se may be a motivation but I thought you something more that this, e.g. that some integrals of this type appeared somewhere else. $\endgroup$
    – Dirk
    Oct 21, 2014 at 9:42

2 Answers 2

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Let $k$ be the degree of $P$. The case $k<2$ is trivial. Let now $k\ge2$. Let us show that then both integrals converge.

Indeed, without loss of generality, the leading coefficient of $P$ is positive. Then, since $k\ge2$, there is some real $a$ such that $P(x)>0$, $P'(x)>0$, and $P''(x)>0$ for $x\ge a$, so that the differentiable functions $P$ and $P'$ are positive and increasing on the interval $[a,\infty)$, and $P$ maps $[a,\infty)$ onto the interval $[b,\infty)$, where $b:=P(a)$. Let $Q\colon[b,\infty)\to[a,\infty)$ be the function inverse to the restriction $P\big|_{[a,\infty)}$ of $P$ on $[a,\infty)$. For simplicity of writing, let $P$ and $P'$ also denote the restriction $P\big|_{[a,\infty)}$ and its derivative. Then the function $\psi:=Q'=1/(P'\circ Q)$ is positive and decreasing, since $P'$ is positive and increasing and $Q$ is increasing. Moreover, $\psi(u)\to0$ as $u\to\infty$, since $P'(x)\to\infty$ as $x\to\infty$ and $Q(u)\to0$ as $u\to\infty$. Now, for any integer $m\ge b/(2\pi)$, we have $$\int_{Q(2\pi m)}^\infty\sin (P(x))\,dx=\int_{2\pi m}^\infty\sin(u)\,\psi(u)\,du =\sum_{j=0}^\infty(-1)^j I_j, $$ where $$I_j:=\int_0^\pi\sin(t)\,\psi(t+2\pi m+\pi j)\,dt. $$ Because $\sin>0$ on $(0,\pi)$, $\psi$ is positive and decreasing, and $\psi(u)\to0$ as $u\to\infty$, it follows that $I_j$ is positive and decreasing in $j$, and $I_j\to0$ as $j\to\infty$. So, the alternating series $\sum_{j=0}^\infty(-1)^j I_j$ converges, and hence so do the integrals $\int_{Q(2\pi m)}^\infty\sin (P(x))\,dx$ and $\int_0^\infty\sin (P(x))\,dx$. Quite similarly, the integral $\int_0^\infty\cos (P(x))\,dx$ converges.

Another way to finish the solution (again for $k\ge2$) is to use integration by parts. Indeed, since $P'$ is positive and increasing on $[a,\infty)$ and $P'(x)\to\infty$ as $x\to\infty$, one has $$i\int_a^\infty e^{iP(x)}\,dx=\int_a^\infty \frac1{P'(x)}\cdot e^{iP(x)}\,iP'(x)\,dx =-\frac{e^{iP(a)}}{P'(a)} -\,\int_a^\infty e^{iP(x)}\,\Big(\frac1{P'(x)}\Big)'\,dx. $$ At that,
$$\int_a^\infty \Big|e^{iP(x)}\,\Big(\frac1{P'(x)}\Big)'\Big|\,dx =-\int_a^\infty \Big(\frac1{P'(x)}\Big)'\,dx=\frac1{P'(a)}<\infty. $$ It follows that the integral $\int_0^\infty e^{iP(x)}\,dx$ is converging, and hence so are the integrals $\int_0^\infty\sin (P(x))\,dx$ and $\int_0^\infty\cos (P(x))\,dx$.

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    $\begingroup$ I have added another way to finish the solution, using integration by parts. $\endgroup$ Nov 13, 2015 at 17:18
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The simplest way (unlike my previous answer) is to note that the convergence is easy to show for $P(x) = x^k$ (by integration by parts). Then, for arbitrary polynomials, you write down the integral from $0$ to $t$ by expanding the integrand in a power series, and integrating term by term. Clearly, the integrals are within a constant factor of replacing $P(x)$ by its leading term for $t$ large.

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  • $\begingroup$ Igor Rivin, your style is very nice. About the problem. As a result of Cauchy--Buniakowsky inequality will be integrals of squares of modules and they are all divergent, is not it? The same as integrals for $k=1$. $\endgroup$
    – Sergei
    Oct 21, 2014 at 18:38
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    $\begingroup$ Igor: I don't understand how "by expanding the integrand in a power series, and integrating term by term" you get that "the integrals are within a constant factor of replacing $P(x)$ by its leading term for $t$ large". $\endgroup$ Nov 13, 2015 at 4:58

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