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 
 A: Hint: For the calculus approach, note that $\tan(\pi/2-x)=\frac{1}{\tan x}$. So split the interval into $0$ to $\pi/4$, and $\pi/4$ to $\pi/2$. For the second half, make the change of variable $x=\pi/2-u$.
When we make the change of variable, we have to wade through a morass of minus signs. One comes  from $dx=-du$; another comes from the fact that the integration bounds will be in the "wrong" order; and a third comes from $\ln(1/t)=-\ln(t)$.  
A: Using Cameron's (now deleted) comment that the integral is $$\int_0^\infty \frac{\log x}{1+x^2}dx$$ split at $x=1$ and use $x\mapsto x^{-1}$.
A: Going from @PeterTamaroff's integral form,
$$\int_0^{\infty} dx \frac{\log{x}}{1+x^2}$$
we may employ a contour integration.  Consider the integral
$$\oint_C dz \frac{\log^2{z}}{1+z^2}$$
where $C$ is a keyhole contour about the positive real axis.  The integral vanishes along the outer and inner circular contours about the origin.  The contour integral is then equal to the integral up and back along the positive real axis.  Recalling that, below the axis, $z=x e^{i 2 \pi}$, we may write the iintegral as
$$\int_0^{\infty} dx \frac{\log^2{x}-(\log{x}+i 2 \pi)^2}{1+x^2} = -i 4 \pi \int_0^{\infty} dx \frac{\log{x}}{1+x^2} +4 \pi^2 \int_0^{\infty}\frac{dx}{1+x^2}$$
This integral is equal to $i 2 \pi$ times the sum of the residues at the poles $z=\pm i $, or $z_1 = e^{i \pi/2}$ and $z_2=e^{i 3 \pi/2}$:
$$i 2 \pi \left ( \frac{-\pi^2/4}{2 i} + \frac{-9 \pi^2/4}{-2 i}\right )= 2 \pi^3$$
Knowing that
$$\int_0^{\infty}\frac{dx}{1+x^2} = \frac{\pi}{2}$$
we have
$$-i 4 \pi \int_0^{\infty} dx \frac{\log{x}}{1+x^2} +2 \pi^3 =2 \pi^3$$
or
$$\int_0^{\infty} dx \frac{\log{x}}{1+x^2} = 0$$
A: As $\int_a^bf(x)dx=\int_a^bf(a+b-x)dx,$
$$I=\int_0^{\frac\pi2}\ln(\tan x) dx=\int_0^{\frac\pi2}\ln\left(\tan \left(\frac\pi2+0-x\right)\right) dx=\int_0^{\frac\pi2}\ln(\cot x) dx$$
$$=\int_0^{\frac\pi2}\ln (\tan x)^{-1} dx=-\int_0^{\frac\pi2}\ln(\tan x) dx=-I$$
A: 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 
$$\frac{\partial F(1,0)}{\partial b}= \int^{\infty}_0 \frac{\ln(x)}{x^2+1}\, \,dx=0 $$
A: \begin{eqnarray*}
\color{#66f}{\Large\int_{0}^{\pi/2}\ln\left(\tan\left(x\right)\right)\,{\rm d}x}
& = &
\int_{-\pi/4}^{\pi/4}\ln\left(\tan\left(x + {\pi \over 4}\right)\right)\,{\rm d}x
\\&=&
\int_{-\pi/4}^{\pi/4}\ \underbrace{\quad%
\ln\left(1 + \tan\left(x\right) \over 1 - \tan\left(x\right)\right)\quad}
_{\displaystyle{\mbox{Odd function of}\ x}}\ {\rm d}x = \color{#66f}{\Large 0}
\end{eqnarray*}
A: We have
$$\log\tan x= \log\sin x-\log\cos x$$
and since sine and cosine are symmetric on $(0,\pi/2)$, with respect to the line $x=\pi/4$, their integrals are equal. That is
$$\int_0^{\pi/2}\log\sin x dx = \int_0^{\pi/2}\log\cos x dx$$ 
A: $$Let$$
$$Tanx-> x$$
$$I=\int_{0}^{\infty }\frac{lnx}{1+x^{2}}dx$$$$substitution x=\frac{1}{x}$$$$I=\int_{0}^{\infty }\frac{-lnx}{1+x^{2}}dx\\I+I=0$$$$I=0$$
