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)π20ln(sinx)dx=π2ln(12)

Explanation:

Suppose it is convergent and put:

S = int_0^(pi/2) ln(sin x)dxS=π20ln(sinx)dx

Substitute t=pi/2-xt=π2x

S=-int_(pi/2)^0ln(sin(pi/2-t))dt = int_0^(pi/2) ln( cos t )dtS=0π2ln(sin(π2t))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=2xπ2 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 ==12π20ln(sin(t))dt+12π20ln(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=12π20ln(sin(t))dt+12π20ln(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)