# How do I evaluate $\int_{0}^{\infty}{\frac{\sqrt{x}\sin(\frac{\cos^3(x)}{x\sqrt x})}{\ln(x+1)}dx}$?

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.

• For the first point, what about using some rough estimate like $\ln (1 + x) \le 1 + x$? Jun 3 at 10:24
• @mattos Thank you! I tried this and the integral in the first point clearly diverges. I will add this to the post Jun 3 at 10:26
• Just out of curiosity : where did you catch this monster ? Jun 3 at 10:38
• 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. Jun 3 at 10:39
• @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 Jun 3 at 10:41

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.

• 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? Jun 3 at 20:42
• That's correct. We get absolute convergence in the original integral iff $(1)$ is absolutely convergent.
– zhw.
Jun 3 at 20:51

It converges, but not absolutely.

# $$0:

For $$0 the integrand is approximately $$\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 $$(2n-\frac12)\pi, the integrand is positive. In the limit $$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, 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.

• 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? Jun 3 at 10:55
• (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. Jun 3 at 10:59
• 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? Jun 3 at 11:07
• 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. Jun 3 at 11:10