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=2∫π20lncosxsin2xdx=2∫10ln(w)(1−w2)32dw(w↦cosx)IBP=2ln(w)w√1−w2|10⏟=0−2∫101√1−w2dw=−2arcsin(1)=−π
Therefore
∫∞0lncos2xx2dx=−π
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