Does $\int_{1}^{\infty}\sin(x\log x)dx $ converge? I'm trying to find out whether $\int_{1}^{\infty}\sin(x\log x)dx $ converges, I know that $\int_{1}^{\infty}\sin(x)dx $ diverges but $\int_{1}^{\infty}\sin(x^2)dx $ converges, more than that, $\int_{1}^{\infty}\sin(x^p)dx $ converges for every $p>0$, so it should be converges in infinity. I'd really love your help with this.
Thanks!

Since $x\log(x)$ is monotonic on $[1,\infty)$, let $f(x)$ be its inverse. That is, for $x\in[0,\infty)$
$$
f(x)\log(f(x))=x\tag{1}
$$
Differentiating implicitly, we get
$$
f'(x)=\frac{1}{\log(f(x))+1}\tag{2}
$$
Then
$$
\begin{align}
\int_1^\infty\sin(x\log(x))\;\mathrm{d}x
&=\int_0^\infty\sin(x)\;\mathrm{d}f(x)\\
&=\int_0^\infty\frac{\sin(x)}{\log(f(x))+1}\mathrm{d}x\tag{3}
\end{align}
$$
Since $\left|\int_0^M\sin(x)\;\mathrm{d}x\right|\le2$ and $\frac{1}{\log(f(x))+1}$ monotonically decreases to $0$, Dirichlet's test (Theorem 17.5) says that $(3)$ converges.
