# How to compute the following integral $I_{\alpha,\beta}$

We have the following identity (see Bateman, H. (1953). Higher Transcendental Functions [Volumes I], p. 25.) $$(*)\quad \Gamma(s)\, \zeta(s,\nu) = \int_{0}^{1} x^{\nu-1} \,(1-x)^{-1} \Bigr(\log 1/x\Big)^{s-1} \, dx; \quad \Re e (s)>1,\Re e (\nu)>0,$$ where $\Gamma(s)$ is the Gamma function and $\zeta(s,\nu)$ is the generalized 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

• I would like compute $I_{\alpha, \beta}$ – Z. Alfata Aug 5 '16 at 14:55
• isn't this just integration by parts? $v'=(1-x)^{-1},u=\log(1/x)^{\beta-1}x^{\alpha}$ – tired Aug 5 '16 at 14:58
• @tired, no, it is not the integration by parts. – Z. Alfata Aug 5 '16 at 16:10

• Thank's. But, we can not apply the identity $(*)$, because $-1<\beta<0$ in my case for the integral $I_{\alpha,\beta}$. – Z. Alfata Aug 5 '16 at 16:07