# Integrate log(sinx) from 0 to pi /2?

Sep 26, 2017

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

#### Explanation:

We use the property ${\int}_{0}^{a} f \left(x\right) \mathrm{dx} = {\int}_{0}^{a} f \left(a - x\right) \mathrm{dx}$

hence we can write $I = {\int}_{0}^{\frac{\pi}{2}} \log \sin x \mathrm{dx} = {\int}_{0}^{\frac{\pi}{2}} \log \sin \left(\frac{\pi}{2} - x\right) \mathrm{dx}$

or $I = {\int}_{0}^{\frac{\pi}{2}} \log \sin x \mathrm{dx} = {\int}_{0}^{\frac{\pi}{2}} \log \cos x \mathrm{dx}$

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

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

= ${\int}_{0}^{\frac{\pi}{2}} \log \sin 2 x \mathrm{dx} - {\int}_{0}^{\frac{\pi}{2}} \log 2 \mathrm{dx}$

= ${\int}_{0}^{\frac{\pi}{2}} \log \sin 2 x \mathrm{dx} - \left(\frac{\pi}{2}\right) \log 2$ .............(A)

Let ${I}_{1} = {\int}_{0}^{\frac{\pi}{2}} \log \sin 2 x \mathrm{dx}$ and $t = 2 x$, then ${I}_{1} = \frac{1}{2} {\int}_{0}^{\pi} \log \sin t \mathrm{dt}$

and using the property ${\int}_{0}^{2 a} f \left(x\right) \mathrm{dx} = 2 {\int}_{0}^{a} f \left(a - x\right) \mathrm{dx}$, if $f \left(2 a - x\right) = f \left(x\right)$ - note that here $\log \sin t = \log \sin \left(\pi - t\right)$ and we get

${I}_{1} = \frac{1}{2} {\int}_{0}^{\pi} \log \sin t \mathrm{dt} = {\int}_{0}^{\frac{\pi}{2}} \log \sin t \mathrm{dt} = I$

Hence (A) becomes $2 I = I - \left(\frac{\pi}{2}\right) \log 2$

or $I = - \left(\frac{\pi}{2}\right) \log 2$