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Consider

$$\int_{0}^{1}{x^{ak}\over 1+x^a}\cdot(-\ln{x})^s\mathrm dx=I\tag1$$ $(a,k,s)\ge 1$

How can we find the closed form for $(1)$?

An attempt:

Applying geometric series to $(1)$ then becomes

$$\int_{0}^{1}(x^{ak}-x^{a(k+1)}+x^{a(k+2)}\cdots)\cdot{(-\ln{x})^s}\mathrm dx\tag2$$

It would be quite long to evaluate $(2)$ when applying integration by parts, so how else do we tackle $(2)$?

I got

$$I=\Gamma(1+s)\left[\sum_{n=0}^{\infty}{(-1)^n\over (an+1)^{1+s}}-\sum_{n=0}^{k-1}{(-1)^n\over (an+1)^{1+s}}\right]$$

  • 1
    since you have asked many such questions now, you should have a feeling about this kind of integrals. what does this feeling tells you? – 2017-02-13
  • 0
    I have reduced the problem to find a closed form for the series $$\Gamma(s+1) \sum_{n=0}^\infty \frac{(-1)^n }{(a(n+k)+1)^{s+1}}$$ but I think this will get not very easy. – 2017-02-13
  • 0
    When I have asked 1001 questions then I will tell you how I feel @Tired – 2017-02-13

1 Answers 1

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$$ \begin{align} I &= \int_{0}^{1}\frac{x^{ak}}{1+x^a}\left(-\log{x}\right)^{s}\,dx \\[4mm] &\qquad \color{red}{-\log{x}=\frac{t}{2a}} \,\Rightarrow\, x=e^{-t/2a},\,\, dx=-\frac{1}{2a}e^{-t/2a},\,\, {\small\int_{0}^{1}}\cdots\,dx=-{\small\int_{0}^{\infty}}\cdots\,dt \\[4mm] &= \frac{1}{(2a)^{s+1}}\int_{0}^{\infty}\frac{t^s\,{\large e}^{-\frac{1+ak}{2a}t}}{1+{\large e}^{-t/2}}\,dt \qquad\qquad\left\{\,{\small\times\frac{1-e^{-t/2}}{1-e^{-t/2}}}\,\right\} \\[4mm] &= \frac{1}{(2a)^{s+1}}\int_{0}^{\infty}\left[\frac{t^s\,{\large e}^{-\frac{1+ak}{2a}t}}{1-{\large e}^{-t}}-\frac{t^s\,{\large e}^{-\frac{1+ak+a}{2a}t}}{1-{\large e}^{-t}}\right]\,dt \qquad\left\{\,{\small\Gamma(s)\zeta(s,\,a)=\int_{0}^{\infty}\frac{t^{s-1}\,e^{-a t}}{1-e^{-t}}\,dt}\,\right\} \\[4mm] &= \color{red}{\frac{\Gamma(s+1)}{(2a)^{s+1}}\left[\zeta\left(s+1,\frac{1+a\,k}{2a}\right)-\zeta\left(s+1,\frac{1+a\,(k+1)}{2a}\right)\right]} \end{align} $$
$\,\qquad\qquad\zeta(s,\,a) \,$ Hurwitz Zeta Function.