It isn't what it looks!
∫ 0 2 π x 2 ( ln ∣ ∣ ∣ 2 sin 2 x ∣ ∣ ∣ ) 2 d x = B A π C
If the equation above holds true for integers A , B and C with A , B coprime positive integers, find A + B + C .
The answer is 63.
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1 solution
Inspiring solution ! (+1) . In other words it's the Log-Sine Integral L 5 2
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It is − L s 5 ( 2 ) ( 2 π ) ; don't forget the minus sign!
@Aditya Narayan Sharma Can you please provide me closed form for the value of L 6 ( 3 ) ( π )
If we consider the function F ( β , ν ) = 2 ν − 1 ∫ 0 π e i β y sin ν − 1 y d y = ν B ( 2 ν + β + 1 , 2 ν − β + 1 ) π e 2 1 i β π = Γ ( 2 ν + β + 1 ) Γ ( 2 ν − β + 1 ) π e 2 1 i β π Γ ( ν ) then G ( ν ) = ∂ β 2 ∂ 2 F ( 0 , ν ) = − 2 ν − 1 ∫ 0 π y 2 sin ν − 1 y d y = − 4 Γ ( 2 ν + 1 ) 2 π Γ ( ν ) [ π 2 + 2 ψ ′ ( 2 ν + 1 ) ] and hence ∫ 0 2 π x 2 ( ln ∣ 2 sin 2 1 x ∣ ) 2 d x = 8 ∫ 0 π y 2 ( ln ∣ 2 sin y ∣ ) 2 d y = − 8 G ′ ′ ( 1 ) = 4 5 1 3 π 5 making the answer 1 3 + 4 5 + 5 = 6 3 .