Why are these two definite integrals equal? How can one prove that, for $0< z<1$, the two integrals
$$\int_0^\infty \frac{u^{z-1}}{1+u}du$$ 
and 
$$\int_0^\infty \frac{u^{-z}}{1+u} du$$
are equal? 
From the integral representation of the beta function
$$B(z,w)=\frac12\int_{0}^\infty \frac{u^{z-1}+u^{w-1}}{(1+u)^{z+w}} du$$
If we replace $w$ with $1-z$, the left hand side equal to $\pi/\sin(\pi z)$ while the right hand side is
$$\frac12\int_{0}^\infty \frac{u^{z-1}+u^{-z}}{(1+u)^{z+w}} du$$
this is the reason of my question. 

If you know, say, that the result of the first integral is $\pi/\sin(\pi z)$, you can see that the second one follows from the first by a mapping $z \mapsto 1-z$, which means the result of the second one is just $\pi/\sin(\pi - \pi z) = \pi/\sin(\pi z)$.
