Suppose it is convergent and put:
#S = int_0^(pi/2) ln(sin x)dx#
Substitute #t=pi/2-x#
#S=-int_(pi/2)^0ln(sin(pi/2-t))dt = int_0^(pi/2) ln( cos t )dt#
Summing the two expressions:
#2S = int_0^(pi/2) ln(sin x)dx + int_0^(pi/2) ln( cos x )dx#
and as the integral is linear:
#2S = int_0^(pi/2) ln(sin x)dx + ln( cos x )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 =#
#int_0^(pi/2) ln(1/2)dx +int_0^(pi/2) ln(sin(2x) )dx =#
# = pi/2ln(1/2) + int_0^(pi/2) ln(sin(2x) )dx#
Let's evaluate this last integral, by splitting in two half intervals and substituting #t=2x# in the first and #t=2x-pi/2# 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 = #
# =1/2int_0^(pi/2) ln(sin(t) )dt+1/2int_0^(pi/2) ln(sin(t+pi/2) )dt =#
# =1/2int_0^(pi/2) ln(sin(t) )dt+1/2int_0^(pi/2) ln(cost )dt = 1/2S+1/2S = S#
Substituting this in the formula above:
#2S = pi/2ln(1/2) + S#
that is:
#S = pi/2ln(1/2)#
