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I am trying to deduce if the following integral converges or diverges. If it converges I would also like to check for absolut convergence:

$$\int_{0}^{\infty}{\frac{\sqrt{x}\sin(\frac{\cos^3(x)}{x\sqrt x})}{\ln(x+1)}dx}$$

Here is what I have tried:

  • We know that $|\frac{\sqrt{x}\sin(\frac{\cos^3(x)}{x\sqrt x})}{\ln(x+1)}| \le \frac{\sqrt{x}}{\ln(x+1)}$, therefore I tried to prove that $\int_{0}^{\infty}\frac{\sqrt{x}}{\ln(x+1)}dx$ converges. However, if we notice that $\ln(x+1) \le x+1$, it is easily deducible that it instead diverges.

  • The integrals $\int_{0}^{\infty}\frac{1}{\ln(x+1)}dx$ and $\int_{0}^{\infty}\sqrt{x}dx$ both diverge. Therefore dividing the initial function by the former and performing a ratio test will propably not work.

  • Since $\int_{0}^{\infty}\frac{1}{\ln(x+1)}dx$ and $\int_{0}^{\infty}\sqrt{x}dx$ both diverge, I tried dividing the initial function by the latter and then performing a ratio test in the hopes that $\lim{\frac{\sin(\frac{\cos^3(x)}{x\sqrt x})}{\ln(x+1)}dx}$ converges. However this limit does not exist, since the $\sin$ component oscillates between negative and positive values indefinitely while $\ln(x+1)$ is always positive.

  • I manipulated the initial function by: $\sin(\frac{\cos^3(x)}{x\sqrt x}) = \sin((\frac{\cos(x)}{\sqrt{x}})^3)$ and tried to find some usefull substitution but no luck there either.

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  • $\begingroup$ For the first point, what about using some rough estimate like $\ln (1 + x) \le 1 + x$? $\endgroup$ Jun 3, 2021 at 10:24
  • $\begingroup$ @mattos Thank you! I tried this and the integral in the first point clearly diverges. I will add this to the post $\endgroup$
    – Stamatis
    Jun 3, 2021 at 10:26
  • $\begingroup$ Just out of curiosity : where did you catch this monster ? $\endgroup$ Jun 3, 2021 at 10:38
  • $\begingroup$ No worries. Just a suggestion, but if you have found the answer then maybe write the answer below and accept it yourself, which then closes the question. $\endgroup$ Jun 3, 2021 at 10:39
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    $\begingroup$ @ClaudeLeibovici I saw it some time ago on this forum and I bookmarked it. However that question was [closed] and I still wanted an answer, so I gave it my best shot and when I failed I asked it myself $\endgroup$
    – Stamatis
    Jun 3, 2021 at 10:41

2 Answers 2

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It converges, but not absolutely.

$0<x\le1$:

For $0<x\ll1$ the integrand is approximately $\frac{\sqrt{x}\sin\left(\frac{\cos^3(x)}{x\sqrt{x}}\right)}{\ln(x+1)}\approx x^{-1/2}\sin(x^{-3/2})$, which can be integrated and converges (but in a very oscillating manner): $$ \int_0^1 x^{-1/2}\sin(x^{-3/2})dx = E_{\frac{1}{3}}(-i)+E_{\frac{1}{3}}(i)+2 \sin (1) \approx 0.394614 $$

$x\gg1$:

For $x\gg1$ we can approximate $\sin\left(\frac{\cos^3(x)}{x\sqrt{x}}\right)\approx\frac{\cos^3(x)}{x\sqrt{x}}$, which simplifies the limit a bit.

Let's look at positive and negative contributions to the integral:

  • For $n\in\mathbb{N}$ and $(2n-\frac12)\pi<x<(2n+\frac12)\pi$, the integrand is positive. In the limit $n\to\infty$ it is $$ \int_{(2n-1/2)\pi}^{(2n+1/2)\pi} dx\frac{\sqrt{x}\sin\left(\frac{\cos^3(x)}{x\sqrt{x}}\right)}{\ln(x+1)} \approx \frac{1}{\ln(1+2n\pi)}\int_{(2n-1/2)\pi}^{(2n+1/2)\pi}dx \sqrt{x}\frac{\cos^3(x)}{x\sqrt{x}} = \frac{1}{\ln(1+2n\pi)} \frac14 \left[ 3 \text{Ci}\left(\frac{1}{2} (4n+3) \pi \right) +\text{Ci}\left(\frac{3}{2} (4n+3) \pi \right) -3 \text{Ci}\left(\frac{1}{2} (4n+1) \pi\right) -\text{Ci}\left(\frac{3}{2} (4n+1) \pi \right) \right] \approx \frac{1}{\ln(1+2n\pi)} \frac{2}{3n\pi} $$ where I've series-expanded the cosine-integral functions for $n\to\infty$.
  • For $n\in\mathbb{N}$ and $(2n+\frac12)\pi<x<(2n+\frac32)\pi$, the integrand is negative. In the limit $n\to\infty$ it is $$ \int_{(2n+1/2)\pi}^{(2n+3/2)\pi}dx \frac{\sqrt{x}\sin\left(\frac{\cos^3(x)}{x\sqrt{x}}\right)}{\ln(x+1)} \approx \frac{1}{\ln(1+(2n+1)\pi)}\int_{(2n+1/2)\pi}^{(2n+3/2)\pi}dx \sqrt{x}\frac{\cos^3(x)}{x\sqrt{x}} \approx \frac{1}{\ln(1+(2n+1)\pi)} \frac{-2}{3n\pi} $$

The asymptotic (large-$x$) contribution to your integral is therefore approximated by the sum $$ \sum_{n=1}^{\infty}\left(\frac{1}{\ln(1+2n\pi)} \frac{2}{3n\pi}-\frac{1}{\ln(1+(2n+1)\pi)} \frac{2}{3n\pi}\right)= \sum_{n=1}^{\infty}\left(\frac{1}{\ln(1+2n\pi)}-\frac{1}{\ln(1+(2n+1)\pi)}\right) \frac{2}{3n\pi}, $$ which converges, but not absolutely.

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  • $\begingroup$ First of all, thank you for the answer! I understand the approach you employ, of analysing the negative/positive contributors to the integral. However I am having trouble understanding the intermediate results specifically this approximation : $\int_{(2n-1/2)\pi}^{(2n+1/2)\pi} \frac{\sqrt{x}\sin\left(\frac{\cos^3(x)}{x\sqrt{x}}\right)}{\ln(x+1)} \approx \frac{1}{\ln(1+2n\pi)}\int_{(2n-1/2)\pi}^{(2n+1/2)\pi} \sqrt{x}\frac{\cos^3(x)}{x\sqrt{x}}$. Where can I find more material on this? $\endgroup$
    – Stamatis
    Jun 3, 2021 at 10:55
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    $\begingroup$ (1) Pulling the log-denominator out of the integral is done because it depends only very slowly on $x$ and is therefore almost a constant over the integration domain. (2) Approximating $\sin(z)\approx z$ is from a Taylor series. $\endgroup$
    – Roman
    Jun 3, 2021 at 10:59
  • $\begingroup$ On (1): So since for very large n, $ x = 2n\pi (+-)\pi/2$ is a very small change, $ln(1+x)$ is approximately constant at $(x0 + x1)/2 = 2n\pi$? If so, how would one rigorously justify this? $\endgroup$
    – Stamatis
    Jun 3, 2021 at 11:07
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    $\begingroup$ Every approximation I've made can be expressed in terms of a series-expansion for large $n$, including correction terms. Rigour would mean summing these correction terms too and checking that they don't diverge either. $\endgroup$
    – Roman
    Jun 3, 2021 at 11:10
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Same main ideas as Roman's answer, but I think we can simplify some things.

Let $f(x)$ denote the integrand. First, the integral over $[0,1]:$ We have $|f(x)|\le \dfrac{x^{1/2}}{\ln(1+x)}.$ Since $\ln(1+x)/x\to 1$ as $x\to 0^+$ (by the definition of $\ln'(1)$ for example), $\ln(1+x) \ge cx$ on $[0,1]$ for some $c>0.$ It follows that $|f(x)|\le (1/c)x^{-1/2}$ on this interval. Hence $\int_0^1|f| <\infty.$ Therefore $\int_0^1 f$ converges.

The integral over $[1,\infty):$ From Taylor we know $\sin u = u +O(u^3)$ for $u\in [-1,1].$ Thus the $\sin$ term in $f(x)$ equals

$$\frac{\cos^3 x}{x^{3/2}} + O(x^{-9/2}),\,\, x\ge 1.$$

Now we can ignore $O(x^{-9/2}),$ as it leads to an absolutely convergent integral. We're left contemplating

$$\tag 1\int_1^\infty \frac{\cos^3 x}{x\ln(1+x)}\, dx.$$

We're set up to use Dirichlet's theorem. First $\int _1^x \cos^3 x\ dx$ is bounded independent of $x.$ To see this note that $\cos^3 x$ is $2\pi$-periodic, and $\int_0^{2\pi}\cos^3 x\,dx =0.$ Second, $1/(x\ln(1+x))$ decreases to $0$ on $[1,\infty).$ By Dirichlet's theorem, $(1)$ converges, and we're done.

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  • $\begingroup$ Thanks! I have have a question though. If I wanted to check absolute convergence, I would only have to check the last integral( (1) in your answer’s notation). Is this correct? $\endgroup$
    – Stamatis
    Jun 3, 2021 at 20:42
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    $\begingroup$ That's correct. We get absolute convergence in the original integral iff $(1)$ is absolutely convergent. $\endgroup$
    – zhw.
    Jun 3, 2021 at 20:51

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