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Combining the identity $(1)$ from [1], I am saying the specialization $a_n=\sqrt{n^3+1}$ and $b_n=i$ (here $i$ denotes the imaginary unit, thus $i^2=-1$) for integers $n\geq 1$, and the explanation of the criterion of Fubini's theorem, see [2] if you need it, I can prove that $$\zeta(3)=\int_0^\infty\sum_{n=1}^\infty\frac{\sin(\sqrt{n^3+1}x)}{\sqrt{n^3+1}}e^{-ix}dx.\tag{A}$$ I believe that such reasoning and calculation is right since our functions $f_n(x)=\frac{\sin(\sqrt{n^3+1}x)}{\sqrt{n^3+1}}e^{-ix}$ satisfy for each $n\geq 1$ that $ \left| f_n(x)\right|\leq\frac{1\cdot\left| e^{-ix}\right|}{\sqrt{n^3+1}}=\frac{1}{\sqrt{n^3+1}} $, and we conclude using the comparison test for series.

Question. I was wondering about questions involving this function $$f(x):=\sum_{n=1}^\infty\frac{\sin(\sqrt{n^3+1}x)}{\sqrt{n^3+1}}$$ defined on $[0,\infty)$ that I know how solve or I don't know how solve those.

I know that using the Weierstrass M-test I can to prove that $f(x)$ is continuous on $[0,\infty)$, but how to prove that there no exists (as I suspect) $$\lim_{x\to\infty }f(x)?$$

Many thanks.

Feel free, if you prefer, add hints for some of previous question, instead of a full answer.

References, both from this Mathematics Stack Exchange:

[1] See the answer by D'Aurizio for Cantarini's lemma, identity $(1)$ from: Find the closed form for $\int_{0}^{\infty}\cos{x}\ln\left({1+e^{-x}\over 1-e^{-x}}\right)dx=\sum_{n=0}^{\infty}{1\over n^2+(n+1)^2}$.

[2] See the second paragraph of the answer by Eldredge: When can a sum and integral be interchanged?

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  • $\begingroup$ There must be something to fix in your question as the definite integral has no dependence on $x$. $\endgroup$ – Jon May 19 '17 at 7:59
  • $\begingroup$ Many thanks @Jon $\endgroup$ – user243301 May 19 '17 at 12:37

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