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

  • 2
    $\begingroup$ The integral seems to diverge at $x \to 1$ (for convergence, $\beta$ must exceed $1$). $\endgroup$ – Noam D. Elkies Aug 5 '16 at 17:40
  • 1
    $\begingroup$ It is interesting that $(*)$ does not depend on $s$. $\endgroup$ – 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 $-1<\beta<0$.

  • $\begingroup$ Thank's, But, $-1<\beta<0$ in my case $\endgroup$ – Z. Alfata Aug 5 '16 at 19:09
  • 2
    $\begingroup$ Ok, I have edited the answer to handle all values of $\beta\neq 1,0,-1,...$. $\endgroup$ – user111 Aug 7 '16 at 12:07
  • $\begingroup$ to show the equality $(*)$, is that you used the technique of analytic continuation for $\Re(\mu)>-1$ in $(*)$? $\endgroup$ – Z. Alfata Sep 18 '16 at 9:33

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.