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

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Answer 1 · 2016-12-02T02:44:16

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

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:

$S = pi/2ln(1/2)$