Study the convergence or divergence of improper integral: $\displaystyle \mathrm{I}=\int\limits_{0}^{+\infty} \dfrac{2\sin(x)-\sin(2x)}{x^{\alpha}. \ln(1+3\sqrt{x})}\mathrm{d}x, \ \alpha \in \mathbb{R}$
My attempt: $\mathrm{I}=\mathrm{I}_1+\mathrm{I}_2$ with $\mathrm{I}_1=\displaystyle \mathrm{I}=\int\limits_{0}^{1} \dfrac{2\sin(x)-\sin(2x)}{x^{\alpha}. \ln(1+3\sqrt{x})}\mathrm{d}x, \ \alpha \in \mathbb{R}$ and $ \mathrm{I}_2=\displaystyle \mathrm{I}=\int\limits_{1}^{+\infty} \dfrac{2\sin(x)-\sin(2x)}{x^{\alpha}. \ln(1+3\sqrt{x})}\mathrm{d}x, \ \alpha \in \mathbb{R}$
When $x\to 0$, $\dfrac{2\sin(x)-\sin(2x)}{x^{\alpha}. \ln(1+3\sqrt{x})} \sim \dfrac{x^3}{3x^\alpha\sqrt{x}} \sim \dfrac{1}{3x^{\alpha-\frac{5}{2}}}$
Then $\mathrm{I}_1 $ will converge when $\alpha -\dfrac{5}{2} <1 $ and diverge when $\alpha -\dfrac{5}{2} \ge 1$ due to the consequence: If $\displaystyle \lim_{x\to \infty} \dfrac{f(x)}{g(x)}=k \in (0,+\infty)$, $\displaystyle \int\limits_{a}^{+\infty} f(x)\mathrm{d}x $ converge then $\displaystyle \int\limits_{a}^{+\infty} g(x)\mathrm{d}x $ will converge, $\displaystyle \int\limits_{a}^{+\infty} f(x)\mathrm{d}x $ diverge then $\displaystyle \int\limits_{a}^{+\infty} g(x)\mathrm{d}x $ will diverge.
But when $x\to +\infty$, I can only show that $\mathrm{I}_2$ will converge when $\alpha$ >1 by use: $$ 0 \le \left| \dfrac{2\sin(x)-\sin(2x)}{x^{\alpha}. \ln(1+3\sqrt{x})}\right| \le \dfrac{3}{\ln4 . x^\alpha } $$
With $\alpha \le 1$, $\dfrac{2\sin(x)-\sin(2x)}{x^{\alpha}. \ln(1+3\sqrt{x})} >0$ when $x \in (2k\pi, (2k+1)\pi) $ and negative when $x \in ((2k-1)\pi, 2k\pi) $ $(k\in\mathbb{Z})$. Moreover with $\alpha <0$, I can not do the same with the case $\alpha>1$. Can somebody help me the case $\alpha \le 1$ when $x\to +\infty$?