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Let a continuous speed function $f:\mathbb{R}^+\to\mathbb{R}^+$ exist such that $ [{\{a,b\}\subseteq\mathbb{R}^+\land a<b}] \Rightarrow f(a)<f(b) $. Then $\int_{0}^{\infty}\sin({x f({x})}) d x$ converges.

Looking at the positive and negative parts in pairs, if the integral starts with a positive chunk, the integral is positive, and if the integral starts with a negative chunk, then the integral is negative. Thus the first chunk is an upper bound if the chunk is positive, and the first chunk is a lower bound if the chunk is negative. The chunks approach zero because $f(x)\to\infty$ as $x\to\infty$, thus the upper and lower bounds approach each other.

My problem with this argument is that even though this argument is obvious, the argument seems non-rigorous.

I need the proof to convince Wolfram staff about a bug report: Integrate[Sin[x*Log[x+1]], {x, 0, Infinity}] is claimed to not converge because it converges too slowly.

Nazgand
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    As an improper Riemann integral, it converges as a consequence of Leibniz's alternating series test. As a Lebesgue integral, you need some additional requirements on $f(x)$ for convergence. I'm not good at reading other people's TeX though, so forgive me if I don't actually look at your file. – Brian Moehring Sep 13 '19 at 23:53
  • I am more concerned about having a correct proof than my proof. Thanks for your helpful comment. – Nazgand Sep 14 '19 at 00:02
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    I edited it so it links to a PDF instead of a TEX file. – Nazgand Sep 14 '19 at 00:08
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    Similar questions: https://math.stackexchange.com/questions/105107/prove-int-0-infty-sin-x2-dx-converges and https://math.stackexchange.com/questions/279540/how-to-show-that-int-0-infty-sinx2-dx-converges and https://math.stackexchange.com/questions/794476/does-int-1-infty-left-sin-leftx2-right-rightdx-converge-or-diverge and https://math.stackexchange.com/questions/1870415/proof-of-the-convergence-of-int-0-infty-sinx4-dx-with-riemann-lebesgue and probably many others. – Gerry Myerson Sep 14 '19 at 04:12
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    I guess you have to assume that $f$ is increasing, unbounded and continuous to have the improper Riemann-integrability of $\sin(x f(x))$. – Jack D'Aurizio Sep 14 '19 at 12:01
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    Anyway $\sin(x\log(x+1))$ is definitely improperly Riemann integrable over $\mathbb{R}^+$, see https://math.stackexchange.com/questions/1184373/about-the-integral-int-0-infty-sinx-log-x-dx – Jack D'Aurizio Sep 14 '19 at 12:02
  • Some comments on formatting: First, drop the use of * to indicate multiplication. That is a Comp Sci thing, not a Math thing. When you use it in math, it just makes the reader wonder what operation you are referring to, since mathematicians examine many, many operations, and often have to use the same notations. Second, it is $\int ..., dx$, never $\int ...,\partial x$. – Paul Sinclair Sep 14 '19 at 14:00
  • @JackD'Aurizio I noticed that not assuming f is continuous allows jumping over all the negative parts of the phase. I was assuming the position of the phase did not change when the speed did. I will modify this and change it to conjecture since my incorrect proof does not apply. – Nazgand Sep 15 '19 at 16:39
  • Looking at the positive and negative parts in pairs, if the integral starts with a positive chunk, the integral is positive, and if the integral starts with a negative chunk, then the integral is negative. Thus the first chunk is an upper bound if the chunk is positive, and the first chunk is a lower bound if the chunk is negative. The chunks approach zero because $f(x)\to\infty$ as $x\to\infty$, thus the upper and lower bounds approach each other.

    My problem with this argument is that even though this argument is obvious, the argument seems non-rigorous.

     – Nazgand Sep 16 '19 at 01:36
  • Have you looked, Nazgand, at all of those links I posted for similar problems, to see whether the ideas there apply to your question? – Gerry Myerson Sep 16 '19 at 02:08
  • Not yet. I took a glance but did not look in detail. – Nazgand Sep 16 '19 at 02:30
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    No worries. Math is eternal. – Gerry Myerson Sep 17 '19 at 13:00

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