Proof that 23&lt;log⁡(2)&lt;710 Positive integrals ∫012x(1−x)21+x2dx=π−3 and ∫01x4(1−x)41+x2dx=227−π prove that 3&lt;π&lt;227 Is there a similar argument for the following log⁡(2) inequality? 23&lt;log⁡(2)&lt;710

Question

Proof that 23&amp;lt;log⁡(2)&amp;lt;710  Positive integrals  ∫012x(1−x)21+x2dx=π−3  and  ∫01x4(1−x)41+x2dx=227−π  prove that  3&amp;lt;π&amp;lt;227  Is there a similar argument for the following log⁡(2) inequality?  23&amp;lt;log⁡(2)&amp;lt;710

Jim Hunt · Accepted Answer

There are positive integrals that relate log(2) to its first four convergents:0,1,23,710.  ∫012x1+x2=log⁡(2)  ∫01(1−x)21+x2dx=1−log⁡(2)  ∫01x2(1−x)21+x2dx  ∫01x4(1−x)21+x2dx=710−log⁡(2)  Therefore,  −∫01x2(1−x)21+x2&amp;lt;0&amp;lt;∫01x4(1−x)21+x2dx  23−log⁡(2)&amp;lt;0&amp;lt;710−log⁡(2)  23&amp;lt;log⁡(2)&amp;lt;710  A similar set is available with denominators (1+x):  ∫0111+xdx=log⁡(2)  ∫01x1+xdx=1−log⁡(2)  12∫01x2(1−x)1+xdx=log⁡(2)−23  12∫01x5(1−x)1+xdx=710−log⁡(2)  and series versions are given by