# Proving improper integral convergence implications

Does convergence of improper integral $$\int_0^{\infty}f(t)dt$$ imply convergence of $$\int_0^{\infty}\frac{f(t)}{1 + (f(t))^2}dt$$ when:

1. $$f(t) \ge 0$$
2. $$f(t)$$ arbitrary

My take on the first task:

Since integral converges we have

$$\lim_{T \rightarrow \infty} \int_0^{T}f(t)dt = M \text{ for some }M \in \mathbb{R}$$

now $$\lim_{T \rightarrow \infty} \int_0^{T}\frac{f(t)}{1 + f(t)^2}f(t)dt = \lim_{T \rightarrow \infty} \int_0^{T}\frac{1}{1 + f(t)^2}f(t)dt \le$$

$$\lim_{T \rightarrow \infty} \int_0^{T}\frac{1}{1 + M^2}f(t)dt = \frac{1}{1 + M^2} \lim_{T \rightarrow \infty} \int_0^{T}f(t)dt$$

How could I prove the same theorem with arbitrary $$f(t)$$?
If $$f\geq 0,$$ then $$\frac{1}{1+(f(t))^2}\leq 1,$$ for any $$t.$$ Hence,
$$\int\limits_0^\infty \frac{f(t)}{1+(f(t))^2}dt\leq \int\limits_0^\infty f(t)dt<\infty,$$ since $$f$$ is integrable by assumption.
Via the same process, $$\int\limits_0^\infty\frac{|f(t)|}{|1+(f(t))^2|}dt\leq \int\limits_0^\infty|f(t)|dt,$$ so if $$f$$ is (absolutely) integrable, then so is the other quantity.