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

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

Notice: \begin{aligned} \frac{d}{dx} \left[x^a(1-x)^{-1}\left(\log \frac{1}{x} \right)^{\beta}\right] &= ax^{a-1}(1-x)^{-1}\left(\log \frac{1}{x} \right)^{\beta} \\ &\quad + x^a(1-x)^{-2}\left(\log \frac{1}{x} \right)^{\beta} \\ &\quad + \beta x^{a-1}(1-x)^{-1}\left(\log \frac{1}{x} \right)^{\beta-1} \end{aligned} and hence \begin{aligned} \int_0^1 x^a(1-x)^{-2}\left(\log \frac{1}{x} \right)^{\beta} \,dx &= \int_0^1 \frac{d}{dx} \left[x^a(1-x)^{-1}\left(\log \frac{1}{x} \right)^{\beta}\right]\, dx \\ &\quad - a\int_0^1 x^{a-1}(1-x)^{-1}\left(\log \frac{1}{x} \right)^{\beta}\, dx \\ &\quad - \beta \int_0^1 x^{a-1}(1-x)^{-1}\left(\log \frac{1}{x} \right)^{\beta-1} \, dx \end{aligned} It seems to me that the right hand side of this last equality is solvable (either by the fundamental theorem of calc, or by your identity above).

  • $\begingroup$ Thank's. But, we can not apply the identity $(*)$, because $-1<\beta<0$ in my case for the integral $I_{\alpha,\beta}$. $\endgroup$ – Z. Alfata Aug 5 '16 at 16:07
  • $\begingroup$ For this parameters your integral is simply divergent $\endgroup$ – tired Aug 5 '16 at 19:04

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