Prove $ \left| \int_{a}^{\infty} \{x\} \frac{1 - \log x}{x^2} dx \right| \le 2 \int_{a}^{\infty} \frac{\log x}{x^2} dx $ It this inequality true for $ a > 1 $?
$$
\left| \int_{a}^{\infty} \{x\} \frac{1 - \log x}{x^2} dx \right|
\le 2 \int_{a}^{\infty} \frac{\log x}{x^2} dx.
$$
The expression $ \{ x \} $ denotes the fractional part of a real number $ x $. So $ \{ x \} = x - \lfloor x \rfloor. $
Is the above inequality true? If true, how can I prove it?
I realize that $ 0 \le \{ x \} \lt 1 $, so it will play a role in proving the inequality but I have been unable to figure out how to do the algebraic manipulation to use this fact to prove the inequality. Any help will be appreciated.
 A: Here's a proof for $a\geq e$:
$$
\begin{align*}
\left|\int_a^\infty \{x\} \frac{1-\log x}{x^2} \, \mathrm{d}x \right| &\leq \int_a^\infty \left|\{x\}\right| \frac{\left|1-\log x\right|}{x^2} \, \mathrm{d}x \\
&\leq \int_a^\infty \frac{1+\left|\log x\right|}{x^2} \, \mathrm{d}x \\
&=\int_a^\infty \frac{1+\log x}{x^2}\, \mathrm{d}x \\
&\leq \int_a^\infty \frac{2\log x}{x^2}\, \mathrm{d}x.
\end{align*}
$$
First inequality is from the triangle inequality for integrals, second from $\{x\}<1$ and the ordinary triangle inequality, third inequality holds for $x\geq 1$ and the final inequality comes from $\log x\geq 1$ for $x\geq e$.
For $a\in \left]1,e\right[$,
$$
\left|\int_a^e \{x\} \frac{1-\log x}{x^2} \, \mathrm{d}x\right|\leq \int_a^e \frac{\left|1-\log x\right|}{x^2} \, \mathrm{d}x=\int_a^e \frac{1-\log x}{x^2} \, \mathrm{d}x=\frac{1}{e}-\frac{\log a}{a}
$$
and
$$
\int_a^e \frac{2\log x}{x^2} \, \mathrm{d}x=\frac{2}{a}-\frac{4}{e}+\frac{2\log a}{a}.
$$
Define a function $f:\left[1,e\right]\to \mathbb{R}$, $f(a)=\frac{2}{a}-\frac{5}{e}+\frac{3\log a}{a}$. Then $f'(a)=\frac{1-3\log a}{a^2}$. Since $f'(a)=0$ iff $x=e^{1/3}$ and $f(e^{1/3}), f(1), f(e)\geq 0$, we have that $f(a)\geq 0$ for all $a\in \left[1,e\right]$ and therefore
$$\left|\int_a^e \{x\} \frac{1-\log x}{x^2} \, \mathrm{d}x\right|\leq\int_a^e \frac{2\log x}{x^2} \, \mathrm{d}x.$$
A similar approach could've been used when proving the inequality for $a\geq e$, but I find it unnecessarily complicated. The reason I brute-forced the case $a<e$ is that once the fractional part is estimated away, one is left with the integrand $\frac{1-\log x}{x^2}$. This, however, is not always smaller than $2\log x$, and so one must somehow account for the entire integration interval.
