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.
 A: 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.
A: 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.
