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Using contour integration, for integer $n\ge 2$, I could obtain the result $$\int_0^{\infty}\sin t^n dt=\sin\bigg(\frac{\pi}{2n}\bigg)\Gamma\bigg(1+\frac{1}{n}\bigg).$$ With the definition $f(x) = \int_x^{x+1}\sin(e^t)dt$ and substitution $e^t=u$, $$e^xf(x) = \cos e^x-e^{-1}\cos e^{x+1}-e^x\int_{e^x}^{e^{x+1}}\frac{\cos u}{u^2}du.$$Because $|\cos u |\lt 1$, it follows that $|\int_{e^x}^{e^{x+1}}\frac{\cos u}{u^2}du|\lt e^{-x}-e^{-x-1}$. Using this in the above equality one has $$e^x|f(x)|\le|\cos e^x| + e^{-1}|\cos e^{x+1}|+ e^x|\int_{e^x}^{e^{x+1}}\frac{\cos u}{u^2}du|$$ $$e^x|f(x)|\lt1+e^{-1}+1-e^{-1}$$ $$\implies e^x|f(x)| \lt 2.$$ Since $$\int_0^{\infty}\sin(e^t)dt=\Big[\sum_{k=0}^{N}f(k)\Big]_{N\rightarrow \infty}$$ and the series $2\sum_{k=0}^{\infty}e^{-k}$ is a majorant of the series in the RHS of above, it follows that $\int_0^{\infty}\sin(e^t)dt$ is convergent. Now consider an integral of the form $$\int_0^{\infty}\sin(p(t))dt$$ where $p(t)$ is a polynomial in $t$ of degree $\ge 2$. It seems likely that such an integral should also converge. My question is does this integral converge, and if it does, then is there a general argument that uses the above facts to show this? Also, what can be said if $p(t)$ were a rational function with no non-negative poles?

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    $\begingroup$ How do you go from a majoration of $e^x f(x)$ to one of $e^x |f(x)|$ (which means you should also have a minoration of $e^x f(x)$) ? $\endgroup$ – Joel Cohen Jan 12 '16 at 8:00
  • $\begingroup$ Oh you mean the inequality $e^x|f(x)| \lt 2$ does not follow unless I prove something like $e^xf(x)\gt -(\cos e^x-e^{-1}\cos e^{x+1}+1-e^{-1})$. $\endgroup$ – vnd Jan 12 '16 at 8:06
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    $\begingroup$ For large $t$, $p(t)$ is essentially $t^n$ so that convergence should be identical. $\endgroup$ – Yves Daoust Jan 12 '16 at 8:39
  • $\begingroup$ @vnd : Yes, that's my point. $\endgroup$ – Joel Cohen Jan 12 '16 at 9:54
  • $\begingroup$ @JoelCohen Thanks for pointing out. I have edited, I guess it is fine now. $\endgroup$ – vnd Jan 12 '16 at 9:58
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Consider the set $\{t>0: p(t)=k\pi \text{ for some integer } k\}$, which is able to be written as an increasing sequence $\{a_1,a_2,\dots,\}$. Now since $p$ has degree $\geq 2$, the integral of $\sin p(t)$ over these intervals $[a_n,a_{n+1}]$ becomes an alternating sequence (eventually, as the highest order term dominates), since we have $$ \lim (a_{n+1}-a_n)=0 $$ i.e. the roots are becoming closer and closer.

Thanks for the idea of the following: ncmathsadist (https://math.stackexchange.com/users/4154/ncmathsadist), Prove: $\int_{0}^{\infty} \sin (x^2) dx$ converges., URL (version: 2012-02-03): https://math.stackexchange.com/q/105167

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