Proof that 23<log⁡(2)<710 Positive integrals ∫012x(1−x)21+x2dx=π−3 and ∫01x4(1−x)41+x2dx=227−π prove that 3<π<227 Is there a similar...

piarepm

piarepm

Answered

2022-01-19

Proof that 23<log(2)<710
Positive integrals
012x(1x)21+x2dx=π3
and
01x4(1x)41+x2dx=227π
prove that
3<π<227
Is there a similar argument for the following log(2) inequality?
23<log(2)<710

Answer & Explanation

Jim Hunt

Jim Hunt

Expert

2022-01-19Added 45 answers

There are positive integrals that relate log(2) to its first four convergents:0,1,23,710.
012x1+x2=log(2)
01(1x)21+x2dx=1log(2)
01x2(1x)21+x2dx
01x4(1x)21+x2dx=710log(2)
Therefore,
01x2(1x)21+x2<0<01x4(1x)21+x2dx
23log(2)<0<710log(2)
23<log(2)<710
A similar set is available with denominators (1+x):
0111+xdx=log(2)
01x1+xdx=1log(2)
1201x2(1x)1+xdx=log(2)23
1201x5(1x)1+xdx=710log(2)
and series versions are given by

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