Is it possible to simplify an improper integral using Taylor's series? How can I prove this procedure is correct?
For example, take $$f(\alpha)=\int_0^{\infty} \left(\arctan\left(\frac{1}{x}\right)\right)^{\alpha}\,dx$$
I want to prove that $f(\alpha)$ converges for $\alpha>1$, then I write $$f(\alpha)=f(\alpha)=\int_0^{c} \left(\arctan\left(\frac{1}{x}\right)\right)^{\alpha}\,dx+\int_c^{\infty} \left(\arctan\left(\frac{1}{x}\right)\right)^{\alpha}\,dx$$ $$= \int_0^{c} \left(\arctan\left(\frac{1}{x}\right)\right)^{\alpha}\,dx+\int_c^{\infty} \left(\frac{1}{x} + O\left(\frac{1}{x^2}\right)\right)^{\alpha}\,dx$$
Of course $$\int_0^{c} \left(\arctan\left(\frac{1}{x}\right)\right)^{\alpha}\,dx$$ is limited for every $c\in \mathbb{R}$ since $\arctan$ is a bounded function.
Finally $$\int_c^{\infty} \left(\frac{1}{x} + O\left(\frac{1}{x^2}\right)\right)^{\alpha}\,dx$$ converges if and only if $\alpha>1$.
Let $t=1/x$. You wanted to show that for small $t$, we have $\arctan t=t+O(t^2)$. Note that by basic stuff about Alternating Series, if $0\lt t \le 1$ the "error" when we truncate the Maclaurin series for $\arctan t$ at $t$ has absolute value less than $t^3/3$.