Considering $$ \int_1^\infty\frac{\arctan(9x)}{1+9x^3}\:dx $$
I am trying to show convergence but looking to use Dirichlet's test and wanted to see if we can do it this way.
Are we supposed to show Acrtan(u) is bounded or THE ABSOLUTE Value of arctan(u) is bounded?
Also, looking for Dirichlet test, not comparision test
Method 1.
One may recall that, $$ |\arctan u|\leq \frac{\pi}2, \quad u \in \mathbb{R}, $$ giving $$ \left|\int_1^\infty\frac{\arctan(9x)}{1+9x^3}\:dx\right|\leq \frac{\pi}2\int_1^\infty\frac1{1+9x^3}\:dx\leq \frac{\pi}{18}\int_1^\infty\frac{dx}{x^3}<\infty, $$ the given integral is thus convergent.
Method 2. Dirichlet's test for integrals.
We may write the given integrand as $$ \frac{\arctan(9x)}{1+9x^3}=\frac{\arctan(9x)}{x^2}\times\frac{x^2}{1+9x^3}. $$ One has $$ \left|\int_a^b\frac{\arctan (9x)}{x^2} \:dx\right|\leq \frac{\pi}{2}\left|\int_a^b\frac{1}{x^2} \:dx\right|\leq \pi, \quad b\geq a\geq1, $$ and one has $$ x \mapsto \frac{x^2}{1+9x^3} $$ decreasing over $[1,\infty)$, tending to $0$ as $x \to \infty$. The given integral is thus convergent.
$\arctan(x)$ is positive and bounded by $\frac{\pi}{2}$ on $\mathbb{R}^+$, hence the integral is trivially convergent.
We may notice that: $$ \int_{1}^{+\infty}\frac{\arctan(9x)}{1+9x^3}\,dx = \int_{0}^{1}\frac{\frac{\pi}{2}-\arctan\frac{x}{9}}{x^2+\frac{9}{x}}\,dx $$ where $\int_{0}^{1}\frac{x}{x^3+9}\,dx $ can be computed from partial fraction decomposition, as well as: $$ I(\alpha) = \int_{0}^{1}\frac{x^2\,dx}{(x^3+9)(\alpha^2+x^2)}.$$ By differentiation under the integral sign, to compute the given integral it is enough to compute $\int_{0}^{1/9}I(\alpha)\,d\alpha$.
