We have the following identity (see Bateman, H. (1953). Higher Transcendental Functions [Volumes I], p. 25.) 
$$(\*)\quad \Gamma(\mu)\, \zeta(\mu,\nu) = \int\_{0}^{1} x^{\nu-1} \,(1-x)^{-1} \Bigr(\log 1/x\Big)^{\mu-1} \, dx; \quad \Re e (\mu)>1,\Re e (\nu)>0,$$
where $\Gamma(\mu)$ is the Gamma function and $\zeta(\mu,\nu)$ is the generalized zeta function ([Hurwitz zeta function](http://mathworld.wolfram.com/HurwitzZetaFunction.html)).


Now, I would like compute the following
$$I\_{\alpha,\beta} = \int\_{0}^{1} x^{\alpha} \,(1-x)^{-2} \Bigr(\log 1/x\Big)^{\beta} \, dx; \quad \alpha>0,\, -1<\beta<0.$$
Thank you in advance



How to compute the following integral $I_{\alpha,\beta}$
Assume first $\beta>1$ so that the integral converges and let
$$f(x)=x^{\alpha}(1-x)^{-1}(-\log x)^{\beta}.$$
Then
$$0=\int\_{0}^{1}df\\=\alpha\int\_{0}^{1}x^{\alpha-1}(1-x)^{-1}(-\log x)^{\beta}dx
+ I\_{\alpha,\beta}-\beta\int\_{0}^{1}x^{\alpha-1}(1-x)^{-1}(-\log x)^{\beta-1}dx,$$
where the last two integrals can be expressed with (\*), so that
$$I\_{\alpha,\beta}=\beta\Gamma(\beta)\zeta(\beta,\alpha)-\alpha\Gamma(\beta+1)\zeta(\beta+1,\alpha)=\Gamma(\beta+1)(\zeta(\beta,\alpha)-\alpha\zeta(\beta+1,\alpha)).
$$
Since this equality holds for all $\beta>1$, $\Gamma(\mu)$ is analytic outside of poles at $\mu=0,-1,-2,...$, and $\zeta(\mu,\nu)$, as a function of $\mu$, is analytic everywhere outside of a pole at $\mu=1$, it follows that the above expression of $I\_{\alpha,\beta}$ is valid everywhere except at $\beta=1,0,-1,...$, in particular it is valid for $-1<\beta<0$.


