After I've read an identity involving an integral related with special functions, I've consider a different integral by trials asking to Wolfram Alpha online calculator
Example For the code int_0^1 (-log u)^s/u^s du the online calculator said that is equal to $$(1-s)^{-s-1}\Gamma(s+1)$$ for $-1<\Re s<1$.
Thus we take here $s=\sigma+it$ the complex variable, and the logarithmic function as you see is the natural logarithm. I've refreshed my knowdledges in complex analysis integration, thus I known the more relevant theorems: Cauchy formula and the Residue Theorem. Also I know representations for the Gamma function, and facts about the theory of such function.
Question. How one can prove previous identity? I ask, can you tell us how prove rigurously that $$\int_0^1\frac{(-\log u)^s}{u^s}du=\frac{\Gamma(s+1)}{(1-s)^{s+1}}$$ holds for $-1<\sigma<1$? Then I can encourage to ask more questions about complex analysis. Thanks in advance.
I don't know if my question is obvious from the definiton of the Gamma function, you can take this approach if such is the case, but I would like to see, if the calculations are feasibles, an answer from complex integration theory.
First assume that $0<s<1$.
By the change of variable $x=-\log u$, $u=e^{-x}$, $du=-e^{-x}dx$, one has $$ \int_0^1\frac{(-\log u)^s}{u^s}du=\int_0^\infty x^s e^{-(1-s)x}dx=\frac1{(1-s)^{s+1}}\int_0^\infty v^s e^{-v}dv $$ the latter integral is a standard integral representation of the Euler gamma function, we have thus obtained $$ \int_0^1\frac{(-\log u)^s}{u^s}du=\frac1{(1-s)^{s+1}}\Gamma(s+1), \quad 0<s<1. \tag1 $$ Since, for $0<u<1$, the integrand $(0,1) \ni s \mapsto \frac{(-\log u)^s}{u^s}$ is holomorphic, since $u \mapsto \frac{(-\log u)^s}{u^s}$ is continuous over $(0,1)$, then the left hand side of $(1)$ is a holomorphic function over $(0,1)$ (holomorphic parameter integral) and one may extend $(1)$ by analytic continuation to $-1<\Re s<1$.