When $L \neq 0$, this limit comparison can be use to prove both absolute convergence when $\mu > 1$ and absolute divergence when $\mu \leqslant 1$.
Suppose $\lim_{x \to \infty} x^\mu f(x) = L$. Then for $\epsilon = |L|/2$ there exists $x_0 > 0$ such that when $x \geqslant x_0> 0$ we have
$$| |x^\mu f(x)| - |L|| \leqslant |x^\mu f(x) - L| \leqslant |L|/2.$$
Whence,
$$-|L|/2 \leqslant |x^\mu f(x)| - |L| \leqslant |L|/2\\ \implies |L|/2 \leqslant |x^\mu f(x)| \leqslant 3|L|/2 \\ \implies (|L|/2)x^{-\mu} \leqslant | f(x)| \leqslant (3|L|/2)x^{-\mu} $$
and the integrals of $|f(x)|$ and $x^{-\mu}$ must converge or diverge together.
The limit comparison test is often presented only for the case where $f(x) \geqslant 0$ for all $x > a$. In that case, if $\lim_{x \to \infty}x^\mu f(x) = L > 0$, then there exists $x_0$ such that when $x \geqslant x_0$ we have
$$0 < \frac{L}{2}x^{-\mu} < f(x) < \frac{3L}{2}x^{-\mu},$$
and $\displaystyle \int_a^\infty f(x) \, dx$ diverges if $\mu \leqslant 1$ (using the left inequality) and converges if if $\mu > 1$ (using the right inequality).
An example of the former case is $f(x) = \sin(1/x)$ - where $f$ is eventually positive. Since $\lim_{x \to \infty}x \sin(1/x) = 1$, then $\displaystyle \int_1^\infty \sin(1/x) \, dx$ diverges.
If $L = 0$, then for any $\epsilon > 0$ there exists $x_0$ such that when $x \geqslant x_0$ we have
$$-\epsilon x^{-\mu} < f(x) < \epsilon x^{-\mu},$$
and we can only conclude that $\displaystyle \int_a^\infty f(x) \, dx$ converges if $\mu > 1$ (using the right inequality). An example is the gamma function
$$\displaystyle \Gamma(s) = \int_0^\infty x^{s-1}e^{-x} \, dx,$$
where proof of convergence is facilitated using the test function $x^{-2}$.
If $L = 0$, then we cannot conclude divergence if $\mu \leqslant 1$ since the left lower bound is negative and the integral could converge to a positive value.
We can prove that sequences of right- or left-hand Riemann sums will converge for a monotone function with a convergent improper integral.
Suppose WLOG $f:(0,1] \to \mathbb{R}$ is nonnegative and decreasing. Suppose further that there is a singularity at $x =0$ but $f$ is Riemann integrable on $[c,1]$ for $c > 0$ and the improper integral is convergent:
$$\lim_{c \to 0+}\int_c^1 f(x) \, dx = \int_0^1 f(x) \, dx.$$
Take a uniform partition $P_n = (0,1/n, 2/n, \ldots, (n-1)/n,1).$ Since $f$ is decreasing we have
$$\frac1{n}f\left(\frac{k}{n}\right) \geqslant \int_{k/n}^{(k+1)/n}f(x) \, dx \geqslant \frac1{n}f\left(\frac{k+1}{n}\right), $$
and summing over $k = 1,2, \ldots, n-1$
$$\frac1{n}\sum_{k=1}^{n-1}f\left(\frac{k}{n}\right) \geqslant \int_{1/n}^{1}f(x) \, dx \geqslant \frac1{n}\sum_{k=2}^nf\left(\frac{k}{n}\right). $$
Hence,
$$ \int_{1/n}^{1}f(x) \, dx +\frac{1}{n}f(1) \leqslant \frac1{n}\sum_{k=1}^{n}f\left(\frac{k}{n}\right) \leqslant \int_{1/n}^{1}f(x) \, dx+ \frac{1}{n}f \left(\frac{1}{n} \right).$$
Note that as $n \to \infty$ we have $f(1) /n \to 0$ and since the improper integral is convergent,
$$\lim_{n \to \infty} \int_{1/n}^{1}f(x) \, dx = \int_0^1 f(x) \, dx, \\ \lim_{n \to \infty}\frac{1}{n}f \left(\frac{1}{n} \right) = 0.$$
The second limit follows from monotonicity and the Cauchy criterion which implies that for any $\epsilon > 0$ and all $n$ sufficiently large
$$0 \leqslant \frac{1}{n}f \left(\frac{1}{n} \right) \leqslant 2\int_{1/2n}^{1/n}f(x) \, dx < \epsilon.$$
By the squeeze theorem we have
$$\lim_{n \to \infty}\frac1{n}\sum_{k=1}^nf\left(\frac{k}{n}\right) = \int_0^1 f(x) \, dx.$$
This proof can be generalized for non-uniform partitions. For oscillatory functions like $g(x) = \sin(1/x)/x$, the failure of the sequence of right-hand Riemann sums to converge is, non-monotonicity notwithstanding, related to non-convergence as $n \to \infty$ of
$$ \frac{1}{n}g \left(\frac{1}{n} \right) = \sin n. $$
This particular case appears to have been covered nicely by @Daniel Fischer in
Improper integrals and right-hand Riemann sums
Best Answer
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.