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

Dec 2, 2016

${\int}_{0}^{\frac{\pi}{2}} \ln \left(\sin x\right) \mathrm{dx} = \frac{\pi}{2} \ln \left(\frac{1}{2}\right)$

#### Explanation:

Suppose it is convergent and put:

$S = {\int}_{0}^{\frac{\pi}{2}} \ln \left(\sin x\right) \mathrm{dx}$

Substitute $t = \frac{\pi}{2} - x$

$S = - {\int}_{\frac{\pi}{2}}^{0} \ln \left(\sin \left(\frac{\pi}{2} - t\right)\right) \mathrm{dt} = {\int}_{0}^{\frac{\pi}{2}} \ln \left(\cos t\right) \mathrm{dt}$

Summing the two expressions:

$2 S = {\int}_{0}^{\frac{\pi}{2}} \ln \left(\sin x\right) \mathrm{dx} + {\int}_{0}^{\frac{\pi}{2}} \ln \left(\cos x\right) \mathrm{dx}$

and as the integral is linear:

$2 S = {\int}_{0}^{\frac{\pi}{2}} \ln \left(\sin x\right) \mathrm{dx} + \ln \left(\cos x\right) \mathrm{dx}$

using the properties of logarithms and the formula for the sine of the double angle:

$2 S = {\int}_{0}^{\frac{\pi}{2}} \ln \left(\sin x \cos x\right) \mathrm{dx} = {\int}_{0}^{\frac{\pi}{2}} \ln \left(\frac{1}{2} \sin \left(2 x\right)\right) \mathrm{dx} =$

${\int}_{0}^{\frac{\pi}{2}} \ln \left(\frac{1}{2}\right) \mathrm{dx} + {\int}_{0}^{\frac{\pi}{2}} \ln \left(\sin \left(2 x\right)\right) \mathrm{dx} =$

$= \frac{\pi}{2} \ln \left(\frac{1}{2}\right) + {\int}_{0}^{\frac{\pi}{2}} \ln \left(\sin \left(2 x\right)\right) \mathrm{dx}$

Let's evaluate this last integral, by splitting in two half intervals and substituting $t = 2 x$ in the first and $t = 2 x - \frac{\pi}{2}$ in the second:

${\int}_{0}^{\frac{\pi}{2}} \ln \left(\sin \left(2 x\right)\right) \mathrm{dx} = {\int}_{0}^{\frac{\pi}{4}} \ln \left(\sin \left(2 x\right)\right) \mathrm{dx} + {\int}_{\frac{\pi}{4}}^{\frac{\pi}{2}} \ln \left(\sin \left(2 x\right)\right) \mathrm{dx} =$

$= \frac{1}{2} {\int}_{0}^{\frac{\pi}{2}} \ln \left(\sin \left(t\right)\right) \mathrm{dt} + \frac{1}{2} {\int}_{0}^{\frac{\pi}{2}} \ln \left(\sin \left(t + \frac{\pi}{2}\right)\right) \mathrm{dt} =$

$= \frac{1}{2} {\int}_{0}^{\frac{\pi}{2}} \ln \left(\sin \left(t\right)\right) \mathrm{dt} + \frac{1}{2} {\int}_{0}^{\frac{\pi}{2}} \ln \left(\cos t\right) \mathrm{dt} = \frac{1}{2} S + \frac{1}{2} S = S$

Substituting this in the formula above:

$2 S = \frac{\pi}{2} \ln \left(\frac{1}{2}\right) + S$

that is:

$S = \frac{\pi}{2} \ln \left(\frac{1}{2}\right)$