The integrals are the following ones:
$$\int_{1}^{\infty}\frac{(\log x)^{\alpha}}{x^{2/3}\arctan(1/x)}dx$$ $$\int_{1}^{\infty}\frac{\sin x}{(x^{2}-1)^{\alpha}}dx$$
I would like to know if my arguments are correct or not: I would appreciate any advice. Thanks in advice!
These are my attempts:
Note: When I write $\sim$, I am saying "which has the same character as" or, if you want, I am using a comparison criteria.
Choosing $k >0$ big enough, $$\int_{1}^{\infty}\frac{(\log x)^{\alpha}}{x^{2/3}\arctan(1/x)}dx\sim\int_{k}^{\infty}\frac{(\log x)^{\alpha}}{x^{2/3}\arctan(1/x)}dx.$$ Then, $\arctan(1/x)\sim 1/x$, and so, $$\int_{k}^{\infty}\frac{(\log x)^{\alpha}}{x^{2/3}\arctan(1/x)}dx\sim\int_{k}^{\infty}x^{1/3}(\log x)^{\alpha}dx.$$ At this point, also, there exists a constant $C>0$ such that $\log x\leq Cx$. Thus, $$\int_{k}^{\infty}x^{1/3}(\log x)^{\alpha}dx\leq\int_{k}^{\infty}x^{1/3}(Cx)^{\alpha}dx=\int_{k}^{\infty}C^{\alpha}x^{1/3+\alpha}dx\sim\int_{k}^{\infty}x^{1/3+\alpha}dx=\int_{k}^{\infty}\frac{1}{x^{-1/3-\alpha}}dx.$$ This last integral is convergent if and only if $-1/3-\alpha>1$, which is not possible since $\alpha>0$.
Hence, the first integral is always divergent.
In order to study the second integral, we must notice that $x=1$ is a problematic point (we divide by $0$), so it is convenient to choose $\varepsilon>0$ and write that $$\int_{1}^{\infty}\frac{\sin x}{(x^{2}-1)^{\alpha}}dx=\int_{1}^{1+\varepsilon}\frac{\sin x}{(x^{2}-1)^{\alpha}}dx+\int_{1+\varepsilon}^{\infty}\frac{\sin x}{(x^{2}-1)^{\alpha}}dx.$$
I will call $(1)$ and $(2)$ each of the previous terms from now on.
In $(1)$, $\sin x\sim x$ and $x^{2}$ is negligible. In consequence, $$\int_{1}^{1+\varepsilon}\frac{\sin x}{(x^{2}-1)^{\alpha}}\sim\int_{1}^{1+\varepsilon}\frac{x}{(-1)^{\alpha}}dx,$$ which is clearly convergent no matter what value $\alpha$ has (we could calculate the integral explicitely and obtain a finite result).
In $(2)$, we can choose $k'>0$ big enough and say that $$\int_{1+\varepsilon}^{\infty}\frac{\sin x}{(x^{2}-1)^{\alpha}}dx\sim\int_{k'}^{\infty}\frac{\sin x}{x^{2\alpha}}dx\leq\int_{k'}^{\infty}\frac{1}{x^{2\alpha}}dx,$$ which is convergent if and only if $2\alpha>1$, which is the same as saying $\alpha>1/2$.
Hence, the second integral is convergent if and only if $\alpha>1/2$.