# Proof that $\int_0^1\frac{(-\log u)^s}{u^s}du=\frac{\Gamma(s+1)}{(1-s)^{s+1}}$ for $|\Re s|<1$

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.

• I believe there's a typo on the integral. It should be $du$. – Gabriel Aug 10 '16 at 8:10
• Since you say you know the gamma function $$\Gamma(z)=\int_0^\infty t^{z-1}e^{-t}dt,$$ one can suggest to start with the change of variable $u=e^{-t}$, reducing your integral to $$\int_0^\infty t^{(s+1)-1}e^{-(1-s)t}dt,$$ which already explains the RHS of the identity you are asking about. – Did Aug 10 '16 at 8:28
• Note that the identity $$\Gamma(z)=\int_{0}^{\infty}t^{z-1}e^{-t}dt$$ holds if $\textrm{Re}(z)>0$, so... – Marco Cantarini Aug 10 '16 at 8:34
• @Did: ...that result valid for $s \in (-1,1)$, followed by an analytic continuation of the R.H.S. to its domain of holomorphy - but I suspect that the OP is not familiar with this. – Alex M. Aug 10 '16 at 8:38
• @AlexM. Actually, for every $s$ such that $-1<\Re s<1$. – Did Aug 10 '16 at 9:13

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 $-1<\Re s<1$.
• Your $\underline{second}$ integral, at the top, already 'shows' that $-1 < \Re\left(\, s\,\right) < 1$ to enforce convergence. Isn't it ?. – Felix Marin Aug 11 '16 at 3:03
• @Felix Marin The condition $0<s<1$ (or $-1<s<1$) is sufficient to ensure the convergence of the second integral. But if you consider $s$ can be a non real complex number then you have to justify how you go from $\int_0^\infty x^s e^{-(1-s)x}dx$ to $\frac1{(1-s)^{s+1}}\int_0^\infty v^s e^{-v}dv$, since a change of variable is not automatically valid here. I wanted to avoid this point. Thanks. – Olivier Oloa Aug 12 '16 at 7:42