How can I show that:
$ \int_b^u \frac{\mathrm dx}{\ln x}\leq \frac{2u}{\ln u}$
where:
$ e^2
?
How can I show that:
$ \int_b^u \frac{\mathrm dx}{\ln x}\leq \frac{2u}{\ln u}$
where:
$ e^2
?
Observe that both the left and right hand side of the inequality are increasing, and that the inequality holds for, say, $u = b$. It will suffice to show that:
$\frac{\mathrm d}{\mathrm du} \left({\int_b^u \frac{\mathrm dx}{\log x}}\right) \le \frac{\mathrm d}{\mathrm du} \left({\frac{2u}{\log u}}\right)$
By the Fundamental Theorem of Calculus and the product rule, we evaluate this condition to:
$\frac1{\log u} \le 2\frac1{\log u} + 2u \frac{(-1)\frac1u}{(\log u)^2} = \frac2{\log u}-\frac2{(\log u)^2}$
Since $e^2 < u$, $2 < \log u$, so that: $\dfrac2{(\log u)^2} < \dfrac1{\log u}$.
Therefore, the desired inequality holds, and so the initial integral estimate holds as well.