# 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.) $$(*)\quad \Gamma(\mu)\, \zeta(\mu,\nu) = \int_{0}^{1} x^{\nu-1} \,(1-x)^{-1} \Bigr(\log 1/x\Big)^{\mu-1} \, dx; \quad \Re e (\mu)>1,\Re e (\nu)>0,$$ where $\Gamma(\mu)$ is the Gamma function and $\zeta(\mu,\nu)$ is the generalized zeta function (Hurwitz zeta function).

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

• The integral seems to diverge at $x \to 1$ (for convergence, $\beta$ must exceed $1$). – Noam D. Elkies Aug 5 '16 at 17:40
• It is interesting that $(*)$ does not depend on $s$. – Gerald Edgar Aug 5 '16 at 17:48

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 $-1<\beta<0$.
• Thank's, But, $-1<\beta<0$ in my case – Z. Alfata Aug 5 '16 at 19:09
• Ok, I have edited the answer to handle all values of $\beta\neq 1,0,-1,...$. – user111 Aug 7 '16 at 12:07
• to show the equality $(*)$, is that you used the technique of analytic continuation for $\Re(\mu)>-1$ in $(*)$? – Z. Alfata Sep 18 '16 at 9:33