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How to show that

$$\int_0^1\!\! \left(\frac1{\log^2\left(1-x\right)}-\frac1{x^2}+\frac1x-\frac1{12}\right) \frac{\mathrm{d}x}x=\frac{\ln{(2\pi)}}{12}-\frac{5}{24}+\frac{1}{2\pi^2}\sum_{n=1}^\infty \frac{\ln{n}}{n^2}$$ ?

I have no any idea.

Thanks in advance.

Lucian
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1 Answers1

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From Glaisher-Kinkelin constant: $$ \frac{\log(2\pi)}{12}-\frac{5}{24}+\frac{1}{2\pi^2}\sum\limits_{n=1}^{\infty}\frac{\log(n)}{n^2} =-\frac{\gamma}{12}-\frac{5}{24}+\log{A} $$ The Integral: $$ \begin{align} I =&\int\limits_{0}^{1}\left[\frac{1}{\log^2(1-x)}-\frac{1}{x^2}+\frac{1}{x}-\frac{1}{12}\right]\frac{dx}{x} \\[2mm] &\quad\color{teal}{\left\{{ \small x=1-e^{-t}\Rightarrow dx=e^{-t}dt \,,\, \int_{0}^{1}dx\mapsto\int_{0}^{\infty}dt }\right\}} \\[2mm] =&\int\limits_{0}^{\infty}\left[\frac{1}{(-x)^2}-\frac{1}{\left(1-e^{-x}\right)^2}+\frac{1}{1-e^{-x}}-\frac{1}{12}\right]\frac{e^{-x}dx}{1-e^{-x}} \\[2mm] =&\int\limits_{0}^{\infty}\left[x^{-2}-\frac{e^{2x}}{\left(e^x-1\right)^2}+\frac{e^x}{e^{x}-1}-\frac{1}{12}\right]\frac{dx}{e^x-1} \\[2mm] &\quad\color{teal}{\{ {\small \frac{e^{2x}}{\left(e^x-1\right)^2} =\frac{e^{2x}-1+1}{\left(e^x-1\right)^2}=\frac{1}{\left(e^x-1\right)^2}+\frac{(e^x-1)(e^x+1)}{\left(e^x-1\right)^2}} } \\ &\color{teal}{ {\qquad\small =\frac{1}{\left(e^x-1\right)^2}+\frac{1}{e^x-1}+\frac{e^x}{e^x-1}} \}} \\[2mm] =&\int\limits_{0}^{\infty}\left[x^{-2}-\frac{1}{\left(e^x-1\right)^2}-\frac{1}{e^{x}-1}-\frac{1}{12}\right]\frac{dx}{e^x-1} \\[2mm] =&\int\limits_{0}^{\infty}\left[\frac{x^{-2}}{e^x-1}-\frac{1}{\left(e^x-1\right)^3}-\frac{1}{\left(e^x-1\right)^2}-\frac{1/12}{e^x-1}\right]dx \\[2mm] &\quad\color{teal}{\left\{{ \small +\frac{\frac{1}{2}-\frac{1}{2}}{\left(e^x-1\right)^2}+\frac{\frac{1}{2}-\frac{1}{2}}{e^x-1}+\frac{1-1}{x^3}+\frac{\frac{1}{2}-\frac{1}{2}}{x^2}+\frac{\frac{1}{12}-\frac{1}{12}}{xe^x} }\right\}} \\[5mm] =&\color{magenta}{-\frac{1}{12}\int\limits_{0}^{\infty}\left[\frac{1}{e^x-1}-\frac{1}{xe^x}\right]dx} \\[1mm] &\color{blue}{+\frac{1}{2}\int\limits_{0}^{\infty}\left[\frac{1}{\left(e^x-1\right)^2}+\frac{1}{e^x-1}-\frac{1}{x^2}\right]dx} \\[1mm] &\color{blue}{-\frac{1}{2}\int\limits_{0}^{\infty}\left[\frac{2}{\left(e^x-1\right)^3}+\frac{3}{\left(e^x-1\right)^2}+\frac{1}{e^x-1}-\frac{2}{x^3}\right]dx} \\[1mm] &\color{red}{+\,\int\limits_{0}^{\infty}\left[\frac{x^{-2}}{e^x-1}-\frac{1}{x^3}+\frac{1/2}{x^2}-\frac{1/12}{xe^x}\right]dx} \\[5mm] =&\color{magenta}{-\frac{1}{12}\lim_{s\to0}\left[\,\Gamma(s+1)\zeta(s+1)-\Gamma(s)\,\right]} \\[1mm] &\color{blue}{+\frac{1}{2}\lim_{s\to0}\left[\,\Gamma(s+1)\zeta(s)\,\right] -\frac{1}{2}\lim_{s\to0}\left[\,\Gamma(s+1)\zeta(s-1)\,\right]} \\[1mm] &\color{red}{+\,\lim_{s\to0}\left[\,\Gamma(s-1)\zeta(s-1)-\Gamma(s)/12\,\right]} \\[2mm] =&\color{magenta}{-\frac{\gamma}{12}}\color{blue}{-\frac{1}{4} +\frac{1}{24}}\color{red}{+\log{A}} \Rightarrow \boxed{\,\, {I=-\frac{\gamma}{12}-\frac{5}{24}+\log{A}} \,\,} \\ \end{align} $$

$ \color{blue}{\underline{\text{The Blue Integrals}\colon}} $ $$ \begin{align} \Gamma(s)\zeta(s) &=\int\limits_{0}^{\infty}\left[\frac{1}{e^x-1}-\frac{1}{x}\right]x^{s-1}dx \,\,\,\colon\,0\lt\Re(s)\lt1 \\[1mm] &{{\small\text{IBP}}\begin{cases}u=\frac{1}{e^x-1}-\frac{1}{x}&\implies du=-\left[\frac{e^x}{\left(e^x-1\right)^2}-\frac{1}{x^2}\right]dx \\ dv=x^{s-1}dx &\implies v=x^s/s\end{cases}} \\[1mm] s\Gamma(s)\zeta(s) &=\int\limits_{0}^{\infty}\left[\frac{e^x{\color{blue}{\small -1+1}}}{\left(e^x-1\right)^2}-\frac{1}{x^2}\right]x^{s}dx \\[1mm] \color{blue}{\Gamma(s+1)\zeta(s)} &\color{blue}{=\int\limits_{0}^{\infty}\left[\frac{1}{\left(e^x-1\right)^2}+\frac{1}{e^x-1}-\frac{1}{x^2}\right]x^{s}dx} \\[1mm] \text{re-use this} &\,\text{result} \\[1mm] \Gamma(s)\zeta(s-1) &=\int\limits_{0}^{\infty}\left[\frac{1}{\left(e^x-1\right)^2}+\frac{1}{e^x-1}-\frac{1}{x^2}\right]x^{s-1}dx \\[1mm] &{{\small\text{IBP}}\left\{ {\small u=\frac{1}{\left(e^x-1\right)^2}+\frac{1}{e^x-1}-\frac{1}{x^2} ,\, dv=x^{s-1}dx} \right\}} \\[1mm] \color{blue}{\Gamma(s+1)\zeta(s-1)} &\color{blue}{=\int\limits_{0}^{\infty}\left[\frac{2}{\left(e^x-1\right)^3}+\frac{3}{\left(e^x-1\right)^2}+\frac{1}{e^x-1}-\frac{2}{x^3}\right]x^{s}dx} \\[1mm] \end{align} $$

$ \color{red}{\underline{\text{The Red Integral}\colon}} $
By the summation identity of zeta function: $$ \displaystyle \Gamma(s-N)\zeta(s-N)=\int_{0}^{\infty}x^{s-N-2}\left[\frac{x}{e^x-1}-\left(\sum_{n=0}^{N}B_{n}\frac{x^n}{n!}\right)\right]dx $$ Hence: $$ \begin{align} &\color{red}{\int\limits_{0}^{\infty}\left[\frac{x^{-2}}{e^x-1}-\frac{1}{x^3}+\frac{1/2}{x^2}-\frac{1/12}{xe^x}\right]dx} \\[1mm] =&\int\limits_{0}^{\infty}\left[x^{\color{red}{0-1}-2}\left(\frac{x}{e^x-1}-1+\frac{x}{2}\right)-\frac{1}{12}\left(\frac{x^{\color{red}{0}-1}}{e^x}\right)\right]dx \\[1mm] =&\color{red}{\lim_{s\to0}\left[\,\Gamma(s-1)\zeta(s-1)-\Gamma(s)/12\,\right]} \\[1mm] =&\lim_{s\to0}\left[\,\Gamma(s-1)\zeta(s-1)-(s-1)\Gamma(s-1)/12\,\right] \\[1mm] =&-\frac{1}{12}\lim_{s\to0}\left[s\Gamma(s-1)\right]+\lim_{s\to0}\left[\Gamma(s-1)\left(\zeta(s-1)+\frac{1}{12}\right)\right] \\[1mm] =&-\frac{1}{12}\lim_{s\to0}\left[\frac{\Gamma(s+1)}{s-1}\right]+\lim_{s\to0}\left[\Gamma(s+1)\frac{\zeta(s-1)+\frac{1}{12}}{s(s-1)}\right] \\[1mm] &{\left\{ {\small\text{L'Hopital's Rule}} \right\}} \\[1mm] =&-\frac{1}{12}\frac{\Gamma(1)}{-1}+\frac{\Gamma(1)\zeta^{\prime}(-1)+\Gamma^{\prime}(1)\left(\zeta(-1)+\frac{1}{12}\right)}{-1} \\[1mm] =&\color{red}{\frac{1}{12}-\zeta^{\prime}(-1)}=\color{red}{\log{A}} \end{align} $$

Hazem Orabi
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