Consider the function $I(z)$ defined through the integral $I(z) = \int_0^\infty \frac{u^{z-1}}{1+u} du$, where z is a complex parameter, what is the maximal choice for the domain of the function and is it analytic on the domain?
for the domain part, I consider u as real constant, so $\frac{u^{z-1}}{1+u} = \frac{1}{u(1+u)} * u^z$ and $u^z$ is defined everywhere so does that mean $I(z)$ is defined everywhere?
for the analytic part, also for $\frac{u^{z-1}}{1+u} = \frac{1}{u(1+u)} * u^z$, $u^z$ is analytic so $\frac{u^{z-1}}{1+u}$ is analytic everywhere，and does that mean I(z) is continuous and $\int_\gamma I(z) dz = \int_\gamma \int_0^\infty \frac{u^{z-1}}{1+u} du dz = \int_0^\infty \int_\gamma \frac{u^{z-1}}{1+u} dz du = \int_0^\infty 0 du = 0 $ for any closed contour so I(z) is analytic by morera's theorem? Thanky you for any help.
First of all, you have to ensure that the given integral converges.
$\int_0^1\frac{u^{z-1}}{1+u}\,du$ converges iff (= if and only if) $\int_0^1 u^{z-1}\,du$ does, which holds iff $\Re z>0$.
Similarly, $\int_1^\infty\frac{u^{z-1}}{1+u}\,du=\int_1^\infty\frac{u^{z-2}}{1+u^{-1}}\,du$ converges iff $\int_1^\infty u^{z-2}\,du$ does, i.e. iff $\Re z<1$.
So, $I(z)$ is defined on the strip $0<\Re z<1$. The analyticity (on this strip) can be shown directly by differentiation under the integral sign (which is easy to justify by an argument of "uniformity" on $\epsilon<\Re z<1-\epsilon$ for any sufficiently small $\epsilon>0$).