Showing $\int_{0}^{1}{x^{2n}-x\over 1+x}\cdot{dx\over \ln{x}}=\ln\left({2\over \pi}\cdot{(2n)!!\over (2n-1)!!}\right)$

Question

Integrate

$$I=\int_{0}^{1}{x^{2n}-x\over 1+x}\cdot{dx\over \ln{x}}=\ln\left({2\over \pi}\cdot{(2n)!!\over (2n-1)!!}\right)\tag1$$

$${x^{2n}-x\over 1+x}=\sum_{k=0}^{\infty}(-1)^k(x^{2n}-x)x^k\tag2$$

Sub $(2)$ into $(1)\rightarrow (3)$

$$I=\sum_{n=0}^{\infty}\int_{0}^{1}(x^{2n+k}-x^{k+1})\cdot{dx\over \ln{x}}\tag3$$ Frullani's theorem

$$\int_{0}^{1}(x^m-x^n)\cdot{dx\over \ln{x}}=\ln{m+1 \over n+1}\tag4$$

Apply $(4)$ to $(3)\rightarrow (5)$

$$I=\sum_{n=0}^{\infty}(-1)^n\ln\left({2n+k+1\over k+2}\right)\tag5$$

$${2\over \pi}={(2n-1)!!\over (2n)!}\prod_{k=0}^{\infty}\left({2n+k+1\over k+2}\right)^{(-1)^k}\tag6$$

Answer 1

Enforce the substitution $x\to e^{-x}$ to write

$$\begin{align} I(n)&=\int_0^1 \frac{x^{2n}-x}{1+x}\frac{1}{\log(x)}\,dx\\\\ &=\int_0^\infty \frac{e^{-x}-e^{-2nx}}{x}\frac{e^{-x}}{1+e^{-x}}\,dx\\\\ &=\sum_{k=0}^\infty (-1)^k \int_0^\infty \frac{e^{-(k+2)x}-e^{-(k+2n+1)x}}{x}\\\\ &=\sum_{k=0}^\infty (-1)^k \log\left(\frac{k+2n+1}{k+2}\right)\\\\ &=\sum_{k=1}^\infty (-1)^{k-1} \log\left(\frac{k+2n}{k+1}\right)\\\\ \end{align}$$

Now, note that we can write the partial sum

$$\begin{align} \sum_{k=1}^{2N} (-1)^{k-1} \log\left(\frac{k+2n}{k+1}\right)&=\sum_{k=1}^N \log\left(\frac{2k-1+2n}{2k}\right)-\sum_{k=1}^N \log\left(\frac{2k+2n}{2k+1}\right)\\\\ &=\sum_{k=1}^N \log\left(\frac{2k-1+2n}{2k+2n}\right)+\sum_{k=1}^N \log\left(\frac{2k+1}{2k}\right)\\\\ &=\sum_{k=n+1}^{n+N} \log\left(\frac{2k-1}{2k}\right)+\sum_{k=1}^N \log\left(\frac{2k+1}{2k}\right)\\\\ &=\sum_{k=1}^{n+N} \log\left(\frac{2k-1}{2k}\right)+\sum_{k=1}^{n+N} \log\left(\frac{2k+1}{2k}\right)\\\\ &-\sum_{k=1}^{n} \log\left(\frac{2k-1}{2k}\right)-\sum_{k=N+1}^{n+N} \log\left(\frac{2k+1}{2k}\right)\\\\ &=\sum_{k=1}^{n+N} \log\left(\frac{(2k-1)(2k+1)}{(2k)(2k)}\right)\\\\ &+\log\left(\frac{(2n)!!}{(2n-1)!!}\right)-\sum_{k=N+1}^{n+N} \log\left(\frac{2k+1}{2k}\right)\\\\ \end{align}$$

Recalling Wallis' Product, we see that

$$\lim_{N\to \infty}\sum_{k=1}^{n+N} \log\left(\frac{(2k-1)(2k+1)}{(2k)(2k)}\right)=-\log(\pi/2)$$

And since $\lim_{N\to \infty}\sum_{k=N+1}^{n+N} \log\left(\frac{2k+1}{2k}\right)=0$, we find

$$I(n)=\log\left(\frac{2}{\pi}\frac{(2n)!!}{(2n-1)!!}\right)$$

as was to be shown!!

---

APPENDIX:

In THIS ANSWER, I evaluated the integral

$$J(n)=\int_0^1 \frac{x^{2n+1}-x}{1+x}\frac{1}{\log(x)}\,dx=\log\left(\frac{(2n+1)!!}{(2n)!!}\right)$$

by making use of the integral evaluated herein.

Answer 2

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} \newcommand{\iff}{\Leftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\, #2 \,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$

\begin{align} &\color{#f00}{\int_{0}^{1}{x^{2n} - x \over 1 + x}\,{\dd x\over \ln\pars{x}}} = -\int_{0}^{1}{x^{2n} - x \over 1 + x}\ \overbrace{\int_{0}^{\infty}x^{y}\,\dd y}^{\ds{-\,{1 \over \ln\pars{x}}}}\ \,\dd x = -\int_{0}^{\infty}\int_{0}^{1}{x^{2n + y} - x^{1 + y} \over 1 + x}\,\dd x\,\dd y \\[3mm] = &\ \int_{0}^{\infty}\pars{\int_{0}^{1}{1 - x^{y + 2n} \over 1 + x}\,\dd x - \int_{0}^{1}{1 - x^{y + 1} \over 1 + x}\,\dd x}\,\dd y\tag{1} \end{align}

---

However, by using the well known digamma $\Psi$ function identity $\ds{\left.\int_{0}^{1}{1 - t^{z - 1} \over 1 - t}\,\dd t \,\right\vert_{\ \Re\pars{z}\ >\ 0} = \Psi\pars{z} + \gamma\quad}$ where $\gamma$ is the Euler-Mascheroni constant: \begin{align} \fbox{$\ds{\int_{0}^{1}{1 - x^{z} \over 1 + x}\,\dd x}$} &= 2\int_{0}^{1}{1 - x^{z} \over 1 - x^{2}}\,\dd x - \int_{0}^{1}{1 - x^{z} \over 1 - x}\,\dd x \\[3mm] & = \int_{0}^{1}{x^{-1/2} - x^{z/2 - 1/2} \over 1 - x}\,\dd x - \int_{0}^{1}{1 - x^{z} \over 1 - x}\,\dd x \\[3mm] & = \int_{0}^{1}{1 - x^{z/2 - 1/2} \over 1 - x}\,\dd x - \int_{0}^{1}{1 - x^{-1/2} \over 1 - x}\,\dd x - \int_{0}^{1}{1 - x^{z} \over 1 - x}\,\dd x \\[3mm] & = \fbox{$\ds{% \Psi\pars{{z \over 2} + \half} - \Psi\pars{\half} - \Psi\pars{z + 1} - \gamma}$} \end{align}

we get, after replacing in $\pars{1}$, \begin{align} &\color{#f00}{\int_{0}^{1}{x^{2n} - x \over 1 + x}\,{\dd x\over \ln\pars{x}}} \\[3mm] = &\ \int_{0}^{\infty}\bracks{\Psi\pars{{y \over 2} + n + \half} - \Psi\pars{y + 2n + 1} - \Psi\pars{{y \over 2} + 1} + \Psi\pars{y + 2}}\,\dd y \end{align}

Since $\ds{\Psi\pars{z}\ \stackrel{\mbox{def.}}{=}\ \totald{\ln\pars{\Gamma\pars{z}}}{z}}$: \begin{align} &\color{#f00}{\int_{0}^{1}{x^{2n} - x \over 1 + x}\,{\dd x\over \ln\pars{x}}} = \left.\ln\pars{\Gamma^{2}\pars{y/2 + n + 1/2}\Gamma\pars{y + 2} \over \Gamma^{2}\pars{y/2 + 1}\Gamma\pars{y + 2n + 1}}\right\vert_{\ 0}^{\ \infty} \\[3mm] = &\ \color{#f00}{% \ln\pars{2^{1 - 2n}\,{\Gamma\pars{2n + 1} \over \Gamma^{2}\pars{n + 1/2}}}} \end{align}

Could you simplify it ?.