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Offset logarithmic integral inequality

Prove that

    \[\int_2^{e+1} \frac{\mathrm{d}t}{\ln t} < e\]

Solution

We have successively:

    \begin{align*} \hspace{-1em}\ln x \leq x-1 &\Rightarrow \ln \frac{1}{x} \leq \frac{1}{x}-1 \\ &\Rightarrow -\ln x \leq \frac{1}{x}-1 \\ &\Rightarrow \ln x \geq 1-\frac{1}{x} \\ &\Rightarrow \frac{1}{\ln x} \leq \frac{1}{1-\frac{1}{x}} \\ &\Rightarrow \int_{2}^{e+1} \frac{\mathrm{d}t}{\ln t} < \int_2^{e+1} \frac{\mathrm{d}t}{1-\frac{1}{t}} \\ &\Rightarrow \mathbf{\int_{2}^{e+1} \frac{\mathrm{d}t}{\ln t} < e} \end{align*}

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