# Integral $\int_0^1 \frac{\ln \ln (1/x)}{1+x^{2p}} dx$…Definite Integral

Calculate $$I_1:=\int_0^1 \frac{\ln \ln (1/x)}{1+x^{2p}} dx, \ p \geq 1.$$ I am trying to solve this integral $I_1$. I know how to solve a related integral $I_2$ $$I_2:=\int_0^1 \frac{\ln \ln (1/x)}{1+x^2} dx=\frac{\pi}{4}\bigg(2\ln 2 +3\ln \pi-4\ln\Gamma\big(\frac{1}{4}\big) \bigg)$$ but I am not sure how to use that result here. In this case I just use the substitution $x=e^{-\xi}$ and than use a series expansion. The result is $$I_2=\int_0^\infty \frac{\xi^s e^{-\xi}}{1+e^{-2\xi}} d\xi=\sum_{n=0}^\infty (-1)^n \frac{\Gamma(s+1)}{(2n+1)^{s+1}}=\Gamma(s+1)L(s+1,\chi_4)$$ where L is the Dirichlet L-Function where $\chi_4$ is the unique non-principal character. This result is further simplified but takes some work. I am interested in the general case above, $I_1$ Thanks

• I believe you made some typos in your subscripts. – David H Mar 8 '14 at 8:10
• There is something similar over here . – Felix Marin Mar 8 '14 at 9:00
• @FelixMarin: There is definitely similar over there. :-) – Lucian Mar 8 '14 at 11:28
• I have an answer that can be helpful. – Felix Marin Mar 8 '14 at 15:08
• @FelixMarin As stated in the post, I have calculated that integral that you are saying looks similar. It is $I_2$ in my post. Thanks anyways. I need the general case as this is most important. – Jeff Faraci Mar 8 '14 at 19:50


\begin{align} &\int_{0}^{1}{\ln\pars{\ln\pars{1/x}} \over 1 + x^{2p}}\,\dd x =\overbrace{\int_{\infty}^{1}{\ln\pars{\ln\pars{x}} \over 1 + x^{-2p}} \,\pars{-\,{\dd x \over x^{2}}}}^{\ds{x\ \to\ {1 \over x}}} =\int_{1}^{\infty}{\ln\pars{\ln\pars{x}}x^{-2} \over 1 + x^{-2p}}\,\dd x \\[3mm]&=\underbrace{\int_{0}^{\infty} {\ln\pars{t}\expo{-2t} \over 1 + \expo{-2pt}}\,\expo{t}\,\dd t} _{\ds{x\ \equiv \expo{t}}} =\lim_{\mu \to 0}\partiald{}{\mu}\int_{0}^{\infty}t^{\mu}\expo{-t}\, {1 \over 1 + \expo{-2pt}}\,\dd t \\[3mm]&=\lim_{\mu \to 0}\partiald{}{\mu}\int_{0}^{\infty}t^{\mu}\expo{-t}\, \sum_{\ell = 0}^{\infty}\pars{-1}^{\ell}\expo{-2\ell pt}\,\dd t =\lim_{\mu \to 0}\partiald{}{\mu}\sum_{\ell = 0}^{\infty} \pars{-1}^{\ell}\int_{0}^{\infty}t^{\mu}\expo{-\pars{2\ell p + 1}t}\,\dd t \\[3mm]&=\lim_{\mu \to 0}\partiald{}{\mu}\sum_{\ell = 0}^{\infty} {\pars{-1}^{\ell} \over \pars{2\ell p + 1}^{\mu + 1}} \underbrace{\int_{0}^{\infty}t^{\mu}\expo{-t}\,\dd t}_{\ds{=\ \Gamma\pars{\mu + 1}}} \end{align} where $\ds{\Gamma\pars{z}}$ is the Gamma Function.

$$\int_{0}^{1}{\ln\pars{\ln\pars{1/x}} \over 1 + x^{2p}}\,\dd x =\lim_{\mu \to 0}\partiald{}{\mu}\bracks{\Gamma\pars{\mu + 1}% \sum_{\ell = 0}^{\infty}{\pars{-1}^{\ell} \over \pars{2\ell p + 1}^{\mu + 1}}} \tag{1}$$

Let's reduce the $\ds{\ell}$-sum in the right hand side: \begin{align} &\sum_{\ell = 0}^{\infty}{\pars{-1}^{\ell} \over \pars{2\ell p + 1}^{\mu + 1}} =\sum_{\ell = 0}^{\infty}\braces{% {1 \over \bracks{2\pars{2\ell}p + 1}^{\mu + 1}}- {1 \over \bracks{2\pars{2\ell + 1}p + 1}^{\mu + 1}}} \\[3mm]&={1 \over \pars{4p}^{\mu + 1}}\sum_{\ell = 0}^{\infty}\braces{% {1 \over \bracks{\ell + 1/\pars{4p}}^{\mu + 1}}- {1 \over \bracks{\ell + 1/2 + 1/\pars{4p}}^{\mu + 1}}} \\[3mm]&={1 \over \pars{4p}^{\mu + 1}}\bracks{% \zeta\pars{\mu + 1,{1 \over 4p}} - \zeta\pars{\mu + 1,\half + {1 \over 4p}}} \end{align} where $\ds{\zeta\pars{s,q}}$ is the Generalizated Zeta Function or/and Hurwitz Zeta Function.

$\pars{1}$ is reduced to: \begin{align} &\!\!\!\!\!\int_{0}^{1}{\ln\pars{\ln\pars{1/x}} \over 1 + x^{2p}}\,\dd x =\lim_{\mu \to 0}\partiald{}{\mu}\braces{% {\Gamma\pars{\mu + 1} \over \pars{4p}^{\mu + 1}}\bracks{% \zeta\pars{\mu + 1,{1 \over 4p}} - \zeta\pars{\mu + 1,\half + {1 \over 4p}}}}\tag{2} \end{align}

Also ( see this page ): \begin{align} &\!\!\!\!\!\!\!\! \zeta\pars{\nu + 1,{1 \over 4p}} - \zeta\pars{\nu + 1,\half + {1 \over 4p}} = \sum_{n = 0}^{\infty}{\pars{-1}^{n} \over n!}\, \bracks{\gamma_{n}\pars{1 \over 4p} - \gamma_{n}\pars{\half + {1 \over 4p}}}\nu^{n} \tag{3} \end{align} where $\ds{\gamma_{n}\pars{a}}$ is a Generalizated Stieltjes Constant.

With the expression $\pars{3}$, $\pars{2}$ is reduced to: \begin{align} &\int_{0}^{1}{\ln\pars{\ln\pars{1/x}} \over 1 + x^{2p}}\,\dd x \\[3mm]&=\sum_{n = 0}^{\infty}{\pars{-1}^{n} \over n!}\, \bracks{\gamma_{n}\pars{1 \over 4p} - \gamma_{n}\pars{\half + {1 \over 4p}}}\ \overbrace{\braces{\lim_{\mu \to 0}\partiald{}{\mu}\bracks{% \Gamma\pars{\mu + 1}\mu^{n} \over \pars{4p}^{\mu + 1}}}} ^{\ds{-\bracks{\gamma + \ln\pars{4p}}\delta_{n,0} + \delta_{n,1} \over 4p}} \\[3mm]&={\gamma + \ln\pars{4p} \over 4p}\,\bracks{% \gamma_{0}\pars{\half + {1 \over 4p}} - \gamma_{0}\pars{1 \over 4p}} + {1 \over 4p}\, \bracks{\gamma_{1}\pars{\half + {1 \over 4p}} - \gamma_{1}\pars{1 \over 4p}} \end{align} where $\ds{\gamma}$ is the Euler-Mascheroni Constant.

According to the Blagouchine paper: $\ds{\gamma_{0}\pars{v} = -\Psi\pars{v}}$ where $\ds{\Psi\pars{v}}$ is the Digamma Function.

Finally, we arrive to this answer main result: \begin{align} &\color{#00f}{\large\int_{0}^{1}{\ln\pars{\ln\pars{1/x}} \over 1 + x^{2p}}\,\dd x} \\[3mm]&=\color{#00f}{{\gamma + \ln\pars{4p} \over 4p}\,\bracks{% \Psi\pars{1 \over 4p} - \Psi\pars{\half + {1 \over 4p}}} + {1 \over 4p}\, \bracks{\gamma_{1}\pars{\half + {1 \over 4p}} - \gamma_{1}\pars{1 \over 4p}}} \end{align} The constants $\ds{\braces{\gamma_{1}\pars{a}}}$ can be calculated for rational values of $a$ by means of a rather cumbersome expression (see formula $\pars{26}$ in Blagouchine paper ). When $\ds{p = 1}$, the results is somehow simple since we can use formula $\pars{11}$ of Blagouchine paper which is valid when $\ds{{1 \over 4p} + \pars{\half + {1 \over 4p}} = 1}$.

Recently, the paper by Professor Blagouchine was published in Journal of Number Theory as he told me via a comment. See the following link: A theorem for the closed-form evaluation of the first generalized Stieltjes constant at rational arguments and some related summations .

• this is excellent. I will work through it and let you know if I have further questions. Very nice solution, thanks Alot!! – Jeff Faraci Mar 9 '14 at 3:56
• Wanted to add, the paper by Blagouchine looks very helpful as well! Thanks – Jeff Faraci Mar 9 '14 at 4:15
• @Jeff That paper is, indeed, wonderful. It's quite fresh: Feb-2014. Thanks. – Felix Marin Mar 9 '14 at 4:39
• @Felix The preprint of my paper which You mentioned here was recently published by the Journal of Number Theory. The journal version of the paper is considerably enlarged and additionally contains numerous summation formulae with the first Stieltjes constant and the Digamma function, an integral representation for $\ln\Gamma(z)$ at rational $z$ and some interesting series for Euler's constant. If You, or someone else, is interested in this journal version, I can send it privately by e-mail. – Iaroslav Blagouchine Nov 29 '14 at 2:11
• @IaroslavBlagouchine Thanks. It would be nice to receive a preprint at felix@fisica.ciens.ucv.ve . I just added an addendum to my answer to give a link to your published paper. Congratulations. – Felix Marin Nov 29 '14 at 19:35

Just for your information, I used a CAS without any success for the general case. However, I obtained some formulas.

For $p=2$, $$\frac{1}{8} \left(-\gamma _1\left(\frac{1}{8}\right)+\gamma _1\left(\frac{5}{8}\right)-\sqrt{2} (\gamma +\log (8)) \left(\pi +2 \log \left(\cot \left(\frac{\pi }{8}\right)\right)\right)\right)$$ For $p=3$, $$\frac{1}{36} \left(-2 \gamma _1\left(\frac{1}{12}\right)+\gamma _1\left(\frac{5}{12}\right)+2 \gamma _1\left(\frac{7}{12}\right)-\gamma _1\left(\frac{11}{12}\right)+12 \sqrt{3} \log (2) \log \left(\sqrt{3}-1\right)+6 \sqrt{3} \log (3) \log \left(\sqrt{3}-1\right)-12 \sqrt{3} \log (2) \log \left(1+\sqrt{3}\right)-6 \sqrt{3} \log (3) \log \left(1+\sqrt{3}\right)-2 \gamma \left(\pi +3 \sqrt{3} \left(\log \left(1+\sqrt{3}\right)-\log \left(\sqrt{3}-1\right)\right)\right)+\pi \left(-3 \log (3)+\log (16)+12 \log (\pi )-16 \log \left(\Gamma \left(\frac{1}{4}\right)\right)\right)\right)$$ For $p=4$,$$\frac{1}{16} \left(-\gamma _1\left(\frac{1}{16}\right)+\gamma _1\left(\frac{9}{16}\right)+16 \log (2) \sin \left(\frac{\pi }{8}\right) \log \left(\sin \left(\frac{3 \pi }{16}\right)\right)-4 \pi \log (2) \csc \left(\frac{\pi }{8}\right)-16 \log (2) \sin \left(\frac{\pi }{8}\right) \log \left(\cos \left(\frac{3 \pi }{16}\right)\right)+16 \log (2) \cos \left(\frac{\pi }{8}\right) \log \left(\tan \left(\frac{\pi }{16}\right)\right)-\gamma \left(\pi \csc \left(\frac{\pi }{8}\right)+4 \left(\sin \left(\frac{\pi }{8}\right) \left(\log \left(\cos \left(\frac{3 \pi }{16}\right)\right)-\log \left(\sin \left(\frac{3 \pi }{16}\right)\right)\right)+\cos \left(\frac{\pi }{8}\right) \log \left(\cot \left(\frac{\pi }{16}\right)\right)\right)\right)\right)$$ In these formulas, $\gamma$ is the Euler constant and $\gamma_1$ is the Stieltjes constant.

• Totally off topic, but I've just noticed that $$n!= \int_0^1(-\ln x)^ndx= \bigg[\frac{d^n}{dk^n}\int_0^1x^{-k}dx\bigg]_{k=0}= \bigg(\frac1{1-k}\bigg)^{(n)}_{k=0}$$ and wanted to share this with you. – Lucian Mar 8 '14 at 11:26
• @Lucian. I knew it was off topic but I just wanted to provide something to the OP. On the other side, I am very happy by your answer and comments. Thanks and cheers. – Claude Leibovici Mar 8 '14 at 14:40
• LOL! No, I meant my comment! :-) – Lucian Mar 8 '14 at 15:01
• @Lucian That's is always a useful trick. Thanks. – Felix Marin Mar 9 '14 at 4:38