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Wednesday, January 26, 2022

Integral of the day XXII

An integral involving the Lobachvesky integral formula

Nnice integral representation of π


Today we show the proof of this integral posted by @integralsbot 0lncos2xx2dx=π
For the proof we will use the Lobachevsky integral formula.

Proof

Recall the Lobachevsky integral formula:

Proposition. Let f(x) a complex-valued, π-periodic function that is absolutely integrable over a single period. Therefore 0sin2xx2f(x)dx=0sinxxf(x)dx=π20f(x)dx
Note that 0lncos2xx2dx=0sin2xx2lncos2xsin2xdx
Since the function f(x)=lncos2xsin2xdx
is π-periodic it follows that I=0lncos2xx2dx=0sin2xx2lncos2xsin2xdx=π20lncos2xsin2xdx
Hence I=π20lncos2xsin2xdx=2π20lncosxsin2xdx=210ln(w)(1w2)32dw(wcosx)IBP=2ln(w)w1w2|10=021011w2dw=2arcsin(1)=π
Therefore 0lncos2xx2dx=π

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