If $\int_a^\infty f(x)dx$ exists (converges), then $\int_a^\infty \frac{f(x)}{x}dx$ exists This is a true or false statement:
If $\displaystyle \int_a^\infty f(x)dx$ exists (converges), then $\displaystyle\int_a^\infty \frac{f(x)}{x}dx$ converges for $a>0$ (and $a\in\mathbb{R^+}$).
In first place I thought this was false, but tried a numerous amount of functions but none of them works as a counterexample. It also confuses me that $a>0$, since I'm not sure if an inegral where I take $a=10^{-50}$ (for instance) it should diverge if it's a "huge" number or not.
It might be true, but I can't prove it. Any hint?
PD: I don't know if working with bounded functions in the interval $[a,1]$ may also work.
 A: This is true if  the improper integral $\displaystyle\int_a^\infty f(x) \, dx$ is conditionally convergent, even if $f \not\in L^1([a,\infty))$ and $f$ is neither nonnegative nor nonpositive.
Note that  $x \mapsto \frac{1}{x}$ is decreasing.  By the second mean value theorem for integrals, for any $c_2 > c_1 > a$  there exists $\xi \in (c_1,c_2)$ such that
$$\int_{c_1}^{c_2}\frac{f(x)}{x} \, dx = \frac{1}{c_1} \int_{c_1}^\xi f(x) \, dx,$$
and
$$\left|\int_{c_1}^{c_2}\frac{f(x)}{x} \, dx\right| = \frac{1}{c_1} \left|\int_{c_1}^\xi f(x) \, dx\right|\leqslant \frac{1}{a} \left|\int_{c_1}^\xi f(x) \, dx\right| $$
Since the improper integral of $f$ converges, it follows by the Cauchy criterion that for any $\epsilon > 0$ there exists $C(\epsilon)> a$ such that for all $\xi > c_1 > C(\epsilon)$,
$$ \left|\int_{c_1}^\xi f(x) \, dx\right|< a\epsilon$$
Thus, for all $c_2> c_1 > C(\epsilon)$, we have
$$\left|\int_{c_1}^{c_2}\frac{f(x)}{x} \, dx\right| < \epsilon,$$
and $\displaystyle\int_a^\infty \frac{f(x)}{x} \, dx$ is convergent.
A: I suppose that $f \in L^1(a, \infty)$. By integration by part :
$$\int_a^\infty f(t) \frac1t dt = \Big[ \int_a^t f(u) du \times \frac1t \Big]_{t = a}^{t = \infty} + \int_a^\infty \int_a^t f(u) du \frac1{t^2} dt $$
Because $t \to |\int_a^t f(u) du| $ converges to a finite limite as $t \to \infty$ it is bounded by $M > 0$.
Thus :
$$\int_a^\infty f(t) \frac1t dt = 0 - 0 + \int_a^\infty \int_a^t f(u) du \frac1{t^2} dt $$
And :
$$| \int_a^t f(u) du \frac1{t^2} | \leq M \frac1{t^2} \in L^1(a, \infty)$$
Hence $\int_a^\infty \int_a^t f(u) du \frac1{t^2} dt$ converges.
I don't know if the result is still true when $f$ isn't in $L^1(a, \infty)$.
