definite integral without using complex line integral

Question

It is well known that the following definite integral is proved by using complex line integral.

$\int_{0}^{\pi} \log(\sin{x}) dx =-\pi\log{2}$

Does anyone know the calculation of above integral without using complex line integral?

Answer 1

Notice, the following properties of definite integral

$$\color{blue}{\int_{0}^{2a}f(x)dx=2\int_{0}^{a}f(x)dx\iff f(2a-x)=f(x)}$$ & $$\color{blue}{\int_{0}^{a}f(x)dx=\int_{0}^{a}f(a-x)dx}$$

Now, let's assume $$I=\int_{0}^{\pi}\log(\sin x)dx$$ $$I=2\int_{0}^{\pi/2}\log(\sin x)dx\tag 1$$ $$I=2\int_{0}^{\pi/2}\log\left(\sin\left(\frac{\pi}{2}- x\right)\right)dx$$ $$I=2\int_{0}^{\pi/2}\log\left(\cos x\right)dx\tag 2$$ Adding (1) & (2), we get

$$I+I=2\int_{0}^{\pi/2}\log(\sin x)dx+2\int_{0}^{\pi/2}\log(\cos x)dx$$ $$2I=2\int_{0}^{\pi/2}(\log(\sin x)+\log (\cos x))dx$$ $$I=\int_{0}^{\pi/2}\log\left(\sin x\cos x\right)dx$$

$$I=\int_{0}^{\pi/2}\log\left(\frac{2\sin x\cos x}{2}\right)dx$$ $$I=\int_{0}^{\pi/2}(\log\left(\sin 2x\right)-\log 2)dx$$ $$I=\int_{0}^{\pi/2}\log\left(\sin 2x\right)dx-\log 2\int_{0}^{\pi/2} dx$$ $$I=\frac{1}{2}\int_{0}^{\pi}\log\left(\sin x\right)dx-\frac{\pi}{2}\log 2$$ From (1), we get $$I=\frac{1}{2}I-\frac{\pi}{2}\log 2$$ $$\frac{1}{2}I=-\frac{\pi}{2}\log 2$$ $$I=-\pi\log 2$$

Hence, we have

$$\bbox[5px, border:2px solid #C0A000]{\color{red}{\int_{0}^{\pi}\log(\sin x)dx=-\pi\log2}}$$

Answer 2

Another (though similar) method:

We know that: $$I=\int _0^{\pi}\ln(\sin(x))dx=2\int_0^{\pi/2}\ln(\sin(x))dx$$ $$=2\int_0^{\pi/2}\ln(\sin(\frac \pi 2 -x)dx=2\int_0^{\pi/2}\ln(\cos(x))dx$$ Let $2u=x$ then $2du=dx$ then: $$I=2\int_0^{\pi/2}\ln(\sin(2u))du$$ $$2\int_0^{\pi/2}\ln(\sin(u))du +2\int_0^{\pi/2}\ln(\cos(u))du+2\ln2\int_0^{\pi/2}du$$ $$I=2I+\pi \ln2$$ $$I=-\pi \ln2$$

Answer 3

$$\int_0^{\pi} \log(\sin(x)) dx= \int_0^{\frac{\pi}{2}} \log(\sin(x)) dx + \int_{\frac{\pi}{2}}^{\pi} \log(\sin(x)) dx$$

$$\int_{\frac{\pi}{2}}^{\pi} \log(\sin(x)) dx = \int_0^{\frac{\pi}{2}} \log\left(\sin\left(x+\frac{\pi}{2}\right)\right) dx = \int_0^{\frac{\pi}{2}}\log(\cos(x))dx$$

So:

$$\begin{split} \int_0^{\pi} \log(\sin(x)) dx&= \int_0^{\frac{\pi}{2}} [\log(\sin(x)) + \log(\cos(x))]dx\\ &= \int_0^{\frac{\pi}{2}} \log(\sin(x)\cos(x))dx\\ &= \int_0^{\frac{\pi}{2}} \log\left(\frac{\sin(2x)}{2}\right) dx\\ &= -\frac{\pi\log(2)}{2} + \int_0^{\frac{\pi}{2}}\log(\sin(2x)dx)\\ &= -\frac{\pi\log(2)}{2} + \frac{1}{2}\int_0^{\pi} \log(\sin(x))dx\\ \end{split}$$

This means that:

$$\int_0^{\pi} \log(\sin(x)) dx - \frac{1}{2}\int_0^{\pi} \log(\sin(x))dx= -\frac{\pi\log(2)}{2}$$

And thus:

$$\int_0^{\pi} \log(\sin(x))dx = -\pi\log(2)$$