piarepm

2022-01-19

Proof that $\frac{2}{3}<\mathrm{log}\left(2\right)<\frac{7}{10}$
Positive integrals
${\int }_{0}^{1}\frac{2x{\left(1-x\right)}^{2}}{1+{x}^{2}}dx=\pi -3$
and
${\int }_{0}^{1}\frac{{x}^{4}{\left(1-x\right)}^{4}}{1+{x}^{2}}dx=\frac{22}{7}-\pi$
prove that
$3<\pi <\frac{22}{7}$
Is there a similar argument for the following $\mathrm{log}\left(2\right)$ inequality?
$\frac{2}{3}<\mathrm{log}\left(2\right)<\frac{7}{10}$

Jim Hunt

Expert

There are positive integrals that relate log(2) to its first four convergents:$0,1,\frac{2}{3},\frac{7}{10}.$
${\int }_{0}^{1}\frac{2x}{1+{x}^{2}}=\mathrm{log}\left(2\right)$
${\int }_{0}^{1}\frac{{\left(1-x\right)}^{2}}{1+{x}^{2}}dx=1-\mathrm{log}\left(2\right)$
${\int }_{0}^{1}\frac{{x}^{2}{\left(1-x\right)}^{2}}{1+{x}^{2}}dx$
${\int }_{0}^{1}\frac{{x}^{4}{\left(1-x\right)}^{2}}{1+{x}^{2}}dx=\frac{7}{10}-\mathrm{log}\left(2\right)$
Therefore,
$-{\int }_{0}^{1}\frac{{x}^{2}{\left(1-x\right)}^{2}}{1+{x}^{2}}<0<{\int }_{0}^{1}\frac{{x}^{4}{\left(1-x\right)}^{2}}{1+{x}^{2}}dx$
$\frac{2}{3}-\mathrm{log}\left(2\right)<0<\frac{7}{10}-\mathrm{log}\left(2\right)$
$\frac{2}{3}<\mathrm{log}\left(2\right)<\frac{7}{10}$
A similar set is available with denominators $\left(1+x\right)$:
${\int }_{0}^{1}\frac{1}{1+x}dx=\mathrm{log}\left(2\right)$
${\int }_{0}^{1}\frac{x}{1+x}dx=1-\mathrm{log}\left(2\right)$
$\frac{1}{2}{\int }_{0}^{1}\frac{{x}^{2}\left(1-x\right)}{1+x}dx=\mathrm{log}\left(2\right)-\frac{2}{3}$
$\frac{1}{2}{\int }_{0}^{1}\frac{{x}^{5}\left(1-x\right)}{1+x}dx=\frac{7}{10}-\mathrm{log}\left(2\right)$
and series versions are given by

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