I have to find for which $\alpha$ values this integral converges:
$$\int_0^{\infty}\frac{\arctan{x}-x}{x^\alpha}dx$$
My first approach was to separate the two components in $\frac{\arctan{x}}{x^\alpha}$ and $\frac{x}{x^\alpha}$ but after seeing this For which $\alpha$ does this integral converge? I tried seeing it under another point of view, $\arctan{x}$ can be rewritten (using Taylor) as $x-\frac{x^3}3$. This would make the integral:
$$-\frac13\int_0^{+\infty}\frac1{x^{\alpha-2}} = -\frac13\frac1{(\alpha-1)x^{\alpha-1}}\bigg|_0^t$$
with $t\in [0,+\infty)$. Then
$$-\frac13\frac1{(\alpha-1)0^{\alpha-1}}+\lim_{t\rightarrow+\infty}-\frac13\frac1{(\alpha-1)t^{\alpha-1}}$$
While the second term can clearly converge for $\alpha\geq 1$ where it would converge to $0$, the first term confuses me a bit. How do I go on from here? Did I get something wrong?
Best Answer
The mistake is coming from your Taylor expansion. When $x\to\infty$, $\arctan(x)\to\frac\pi2$ which is bounded, but your Taylor expansion $x-\frac13x^3\to-\infty$. Hence, you need to discuss them separately. For example,
$$\int_0^{\infty}\frac{\arctan{x}-x}{x^\alpha}dx=\int_0^1\frac{\arctan{x}-x}{x^\alpha}dx+\int_1^{\infty}\frac{\arctan{x}-x}{x^\alpha}dx$$
For $\int_0^1\frac{\arctan{x}-x}{x^\alpha}dx$, you can use Taylor expansion, and discuss the convergence/divergence around $x=0$.
$$\int_0^\epsilon\frac{\arctan{x}-x}{x^\alpha}dx\approx -\frac13\int_0^\epsilon\frac{x^3}{x^\alpha}dx\Longrightarrow \alpha<4$$
For $\int_1^{\infty}\frac{\arctan{x}-x}{x^\alpha}dx$, you use the fact that $\arctan(x)$ is bounded by $\frac\pi2$, and discuss the convergence/divergence when $x\to\infty$.
$$\int_1^{\infty}\frac{\arctan{x}-x}{x^\alpha}dx= \int_1^{\infty}\frac{\arctan{x}}{x^\alpha}dx- \int_1^{\infty}\frac{1}{x^{\alpha-1}}dx$$
The second term gives $\alpha>2$. Next, for the first term, we have
$$\int_1^{\infty}\frac{\arctan{x}}{x^\alpha}dx\le \frac\pi2\int_1^{\infty}\frac{1}{x^\alpha}dx\Longrightarrow \alpha>1$$
hence, the integral converges when
$$2<\alpha<4$$