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 $\pi/\sin(\pi - \pi z) = \pi/\sin(\pi z)$.
$$I=\int_{0}^{\infty} \frac{u^{z-1} du}{1+u}$$ Let $u=1/v \implies du=-dv/v^2$, then $$I=\int_{\infty}^{0} \frac{v^{-z+1}(-dv/v^2)}{1+1/v}= \int_{0}^{\infty} \frac{v^{-z} dv}{1+v}.$$
Try a change of variable in the second integral: $x=\frac 1u$. $dx=-\frac 1{u^2}du$. $x(u=0)=\infty$ and $x(u=\infty)=0$. Then: $$\int_0^\infty\frac{u^{-z}}{1+u}du=-\int_\infty^0\frac{x^z}{1+1/x}\frac{1}{x^2}dx=\int_0^\infty\frac{x^{z-1}}{x+1}dx$$
