How do you prove that the integral of ln(sin(x)) on the interval [0, pi/2] is convergent?

1 Answer
Dec 2, 2016

#int_0^(pi/2) ln(sin x)dx = pi/2ln(1/2)#

Explanation:

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)#