Suppose it is convergent and put:
S = int_0^(pi/2) ln(sin x)dxS=∫π20ln(sinx)dx
Substitute t=pi/2-xt=π2−x
S=-int_(pi/2)^0ln(sin(pi/2-t))dt = int_0^(pi/2) ln( cos t )dtS=−∫0π2ln(sin(π2−t))dt=∫π20ln(cost)dt
Summing the two expressions:
2S = int_0^(pi/2) ln(sin x)dx + int_0^(pi/2) ln( cos x )dx2S=∫π20ln(sinx)dx+∫π20ln(cosx)dx
and as the integral is linear:
2S = int_0^(pi/2) ln(sin x)dx + ln( cos x )dx2S=∫π20ln(sinx)dx+ln(cosx)dx
using the properties of logarithms and the formula for the sine of the double angle:
2S = int_0^(pi/2) ln(sin x cos x )dx = int_0^(pi/2) ln(1/2sin(2x) )dx =2S=∫π20ln(sinxcosx)dx=∫π20ln(12sin(2x))dx=
int_0^(pi/2) ln(1/2)dx +int_0^(pi/2) ln(sin(2x) )dx =∫π20ln(12)dx+∫π20ln(sin(2x))dx=
= pi/2ln(1/2) + int_0^(pi/2) ln(sin(2x) )dx=π2ln(12)+∫π20ln(sin(2x))dx
Let's evaluate this last integral, by splitting in two half intervals and substituting t=2xt=2x in the first and t=2x-pi/2t=2x−π2 in the second:
int_0^(pi/2) ln(sin(2x) )dx =int_0^(pi/4) ln(sin(2x) )dx+int_(pi/4)^(pi/2) ln(sin(2x) )dx = ∫π20ln(sin(2x))dx=∫π40ln(sin(2x))dx+∫π2π4ln(sin(2x))dx=
=1/2int_0^(pi/2) ln(sin(t) )dt+1/2int_0^(pi/2) ln(sin(t+pi/2) )dt ==12∫π20ln(sin(t))dt+12∫π20ln(sin(t+π2))dt=
=1/2int_0^(pi/2) ln(sin(t) )dt+1/2int_0^(pi/2) ln(cost )dt = 1/2S+1/2S = S=12∫π20ln(sin(t))dt+12∫π20ln(cost)dt=12S+12S=S
Substituting this in the formula above:
2S = pi/2ln(1/2) + S2S=π2ln(12)+S
that is:
S = pi/2ln(1/2)S=π2ln(12)