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. 

$$I=\int_{0}^{\infty} \frac{u^{z-1} du}{1+u}$$
Let $u=1/v \implies du=-dv/v^2$, then
$$I=\int_{\infty}^{0} \frac{v^{-z+1}(-dv/v^2)}{1+1/v}= \int_{0}^{\infty} \frac{v^{-z} dv}{1+v}.$$
