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:
- $f(t) \ge 0$
- $f(t)$ 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 $$ \lim_{T \rightarrow \infty} \int_0^{T}\frac{f(t)}{1 + f(t)^2}f(t)dt = \lim_{T \rightarrow \infty} \int_0^{T}\frac{1}{1 + f(t)^2}f(t)dt \le $$
$$ \lim_{T \rightarrow \infty} \int_0^{T}\frac{1}{1 + M^2}f(t)dt = \frac{1}{1 + M^2} \lim_{T \rightarrow \infty} \int_0^{T}f(t)dt $$
I am not sure about this though. Is that a correct proof?
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.