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How can I show that:

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

where:

$ e^2

?

1 Answers 1

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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!2012-10-31