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I'm interested in such a simple question. Well, it was simple until I tried to work with it it quite strictly.

Let's consider an improper integral $\int \limits_{1}^{+\infty} f(x)dx$ and let's assume that:

  1. $f(x) \to 0, x \to +\infty$;
  2. $\int \limits_{1}^{+\infty} f(x)dx$ converges (notice that the condition $f(x)\geq 0 \forall x \in [1, +\infty)$ is not required).

Is it true that if we apply a "good" function $\varphi$ to $f(x)$, then $\int \limits_{1}^{+\infty} \varphi(f(x))dx$ will still converge? By "good" I mean such $\varphi(x)$ that, let's say, $1000$ times differentiable in some neighbourhood of $0$ and $\varphi(x) \sim x, x \to 0 $.

For example, is it true that taking into account the conditions above integrals like $\int \limits_{1}^{+\infty} \sin(f(x))dx$ or $\int \limits_{1}^{+\infty} \arctan(f(x))dx$ will still converge?

Anton
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    Consider $\varphi(x) = x + x^2$. – Dermot Craddock Jun 04 '25 at 07:41
  • well, that's a good point. But what if I additionaly require that $\varphi(x)$ is an odd function? – Anton Jun 04 '25 at 07:46
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    $\varphi(x)=x+x^3$? – Lorago Jun 04 '25 at 07:49
  • And I don't see why that $\varphi(x)$ gives a counterexample to my statement – Anton Jun 04 '25 at 07:51
  • I mean, what $f(x)$ should I consider for the integral of $\varphi(f(x))$ to diverge – Anton Jun 04 '25 at 07:52
  • $f^2$ or $f^2$ need not to be in $L^1$: Consider $f(x)=\sum_{n=1}^{+\infty}\sqrt{n}\mathrm{1}_{(n,n+1/n^2)}(x)$. – Nicolas Jun 04 '25 at 08:08
  • There are examples on this site for series $\sum a_n$ which are convergent, but $\sum a_n^3$ diverges. I assume that similar examples can be found for improper integrals. – Martin R Jun 04 '25 at 08:14
  • See here. Although that deals with series, the conclusion also holds for integrals. – Dermot Craddock Jun 04 '25 at 08:22
  • This is what I meant. – Martin R Jun 04 '25 at 08:31
  • Apparently I was wrong, and such a common statement is not true. However, my initial problem was to determine whether the integral $\int \limits_{1}^{+\infty} \arctan(\frac{\cos x}{\sqrt[3]{x^2}})dx$ converges or not. But I'm still sure it converges, however, I cannot prove it – Anton Jun 04 '25 at 08:38
  • That converges. For $\lvert u\rvert \leqslant 1$, we have $\lvert u - \arctan u\rvert \leqslant c\lvert u\rvert^3$ with a constant $c$ I'm too lazy to determine. And $\lvert \frac{\cos x}{\sqrt[3]{x}}\rvert^3 = \frac{\lvert \cos x\rvert^3}{x^2} \leqslant \frac{1}{x^2}$ is absolutely integrable. – Dermot Craddock Jun 04 '25 at 09:04
  • Do you mean it converges even absolutely? – Anton Jun 04 '25 at 09:13
  • No, not that. The integral $$\int_1^{\infty} \arctan\biggl(\frac{\cos x}{\sqrt[3]{x^2}}\biggr) -\frac{\cos x}{\sqrt[3]{x^2}},dx$$ converges absolutely, and $\int_1^{\infty} \frac{\cos x}{\sqrt[3]{x^2}},dx$ converges only conditionally. – Dermot Craddock Jun 04 '25 at 09:37

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