Evaluate : int_0^(pi/2) ln^2(cos^2 x)dx

Rene Nicholson 2022-10-20 Answered
Evaluate :
0 π 2 ln 2 ( cos 2 x ) d x
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Answers (2)

Steinherrjm
Answered 2022-10-21 Author has 12 answers
I find a way to get the number using gamma functions, nothing is rigorous.
Consider the integral I ( β ) = π 2 π 2 ( cos x ) β d x. We know:
2 d 2 d β 2 I ( β ) | β = 0 = 4 0 π 2 ln 2 ( cos x ) d x
is the integral we want. Introduce u = 1 + sin x 2 , we have:
I ( β ) = π 2 π 2 ( cos x ) β 1 d sin x = 0 1 ( 4 u ( 1 u ) ) β 1 2 d ( 2 u ) = 2 β 0 1 u β + 1 2 1 ( 1 u ) β + 1 2 1 d u = 2 β Γ ( β + 1 2 ) 2 Γ ( β + 1 )
Using the taylor expansion of various terms at β = 0
2 β = 1 + ln ( 2 ) β + ln 2 2 2 β 2 + . . . Γ ( β + 1 2 ) = π ( 1 γ + 2 ln 2 2 β + π 2 + 2 ( γ + 2 ln 2 ) 2 16 β 2 + . . . ) Γ ( β + 1 ) = 1 γ β + 6 γ 2 + π 2 12 β 2 + . . .
We get:
I ( β ) = π ( 1 ln ( 2 ) β + π 2 + 12 ln 2 2 24 β 2 + . . . ) 2 d 2 d β 2 I ( β ) | β = 0 = 2 π ( π 2 12 + ln 2 2 )
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Jaylyn Horne
Answered 2022-10-22 Author has 5 answers
0 π 2 sin 2 m 1 ( x ) cos 2 n 1 ( x ) d x = B ( m , n ) 0 π 2 sin 2 m 1 ( x ) ( cos 2 x ) 2 n 1 2 d x = B ( m , n )
On differentiating it twice w.r.t. to n and taking m = 1 2 and n = 1 2 , we get
0 π 2 ( ln ( cos 2 x ) d x ) 2 = 1 2 ( Γ ( 1 2 ) ) 2 1 { ( ψ ( 1 2 ) ψ ( 1 ) ) 2 + ψ ( 1 2 ) ψ ( 1 ) }
0 π 2 ( ln ( cos 2 x ) d x ) 2 = π 3 6 + 2 π ( ln 2 ) 2
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