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

Question

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

Answer 1

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle #1 \right\rangle} \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace} \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left( #1 \right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ $\ds{\int_{0}^{1}{\ln\pars{\ln\pars{1/x}} \over 1 + x^{2p}}\,\dd x\,,\qquad p \geq 1:\ {\large ?}}$

\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 $\ds{{1 \over 4p} + \pars{\half + {1 \over 4p}} = 1}$.

ADDENDUM

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 .

Answer 2

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.