# Showing that $\int_b^u \frac{\mathrm dx}{\ln x}\leq \frac{2u}{\ln u},e^2<b<u$

How can I show that:

$$\int_b^u \frac{\mathrm dx}{\ln x}\leq \frac{2u}{\ln u}$$

where:

$$e^2<b<u$$

?

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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.

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Thank you very much Lord_Farin! –  Chon Oct 31 '12 at 20:00