show that $\int_{0}^{\pi/2}\ln(\tan x)dx=0$ show that $$\int_{0}^{\pi/2}\ln(\tan x)dx=0$$
using two ways The first with real analysis and the second with contour integration 

Consider the following general formula 
$$\tag{1}F(a,b) = \int^{\infty}_0 \frac{x^b}{x^2+a^2}\, \,dx $$
$$F(a,b) = \frac{1}{a^2}\,\int^{\infty}_0 \frac{x^b}{\frac{x^2}{a^2}+1} \, dx$$
Let $\frac{x^2}{a^2} = t$
$$F(a,b) = \frac{a^b}{2|a|} \,\int^{\infty}_0 \frac{t^{\frac{b-1}{2}}}{t+1} \, dt = \frac{a^b \, \pi }{2|a| \sin\left( \pi \frac{b+1}{2}\right)}$$
where I used the beta function .
$$F(a,b)=\frac{a^b \, \pi }{2|a| \sin\left( \pi \frac{b+1}{2}\right)}$$
Differentiate wrt to $b$
$$\frac{\partial F(a,b)}{\partial b} = \frac{a^b \, \pi  \, \ln|a| }{2|a| \sin\left( \pi \frac{b+1}{2}\right)} - \frac{a^b \, \pi^2 }{4|a| } \cot^2 \left(\pi \frac{b+1}{2}\right)$$
$$\frac{\partial F(a,0)}{\partial b}= \int^{\infty}_0 \frac{\ln(x)}{x^2+a^2}\, \,dx=\frac{ \, \pi  \, \ln|a| }{2|a|} $$
In particular 
$$\frac{\partial F(1,0)}{\partial b}= \int^{\infty}_0 \frac{\ln(x)}{x^2+1}\, \,dx=0 $$
