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

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

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.

  • $\begingroup$ What happens when f(t) is arbitrary? $\endgroup$ – math_beginner Jun 8 at 19:04
  • $\begingroup$ @math_beginner I've edited the post! $\endgroup$ – cmk Jun 8 at 19:06

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