How may we show that $\int_{0}^{1}{1-x\over 1+x}\cdot{2k+3+x^2\over 1+x^2}\cdot{\mathrm dx\over \ln x}=-\ln(2^k\pi)?$

Question

A simple closed form

$$\int_{0}^{1}{1-x\over 1+x}\cdot{2k+3+x^2\over 1+x^2}\cdot{\mathrm dx\over \ln x}=-\ln(2^k\pi)\tag1$$ Where $k$ is a real number

Making an attempt:

This integral is too difficult to even try to make an attempt.

Anyway let's try $x=e^y\implies dx=e^ydy$, then $(1)$ becomes

$$-\int_{0}^{\infty}{1-e^{-y}\over 1+e^{-y}}\cdot{2k+3+e^{-2y}\over 1+e^{-2y}}\cdot{e^{-y}\over y}\mathrm dy\tag2$$

Using hyperbolic identities

$$-\int_{0}^{\infty}\tanh\left({y\over 2}\right)\cdot{2k+3+e^{-2y}\over 1+e^{-2y}}\cdot{e^{-y}\over y}\mathrm dy\tag3$$

$$-\int_{0}^{\infty}\tanh\left({y\over 2}\right)\cdot\left[(k+1)e^{2y}sech{y}+1\right]\cdot{e^{-y}\over y}\mathrm dy\tag4$$

Another try: $x=\tan y\implies dx=\sec^2y dy$, then $(1)$ becomes

$$\int_{0}^{\pi/4}\tan\left({\pi\over 4}-y\right)\cdot{2k+2+\sec^2y\over \ln\tan y}\mathrm dy\tag5$$

Still going no where!

How can we prove $(1)?$

Answer 1

PRIMER:

In THIS ANSWER, I showed using the Wallis Product for $\pi$, the

$$\sum_{n=1}^\infty (-1)^{n-1}\log\left(\frac{n+1}{n}\right)=\log\left(\prod_{n=1}^\infty\frac{2n}{2n+1}\frac{2n}{2n-1}\right)=\log(\pi/2) \tag 1$$

We will use $(1)$ in that which follows.

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We first observe that $\frac{x-1}{\log(x)}=\int_0^1 x^t\,dt$. Therefore, we can write

$$\begin{align} \int_0^1 \frac{1-x}{1+x}\frac{x^2+(2k+3)}{x^2+1}\frac{1}{\log(x)}\,dx&=-\int_0^1\int_0^1 \frac{x^t}{1+x}\frac{x^2+(2k+3)}{x^2+1}\,dt\,dx\\\\ &=-\int_0^1 \int_0^1 \frac{x^t}{1+x}\left(1+\frac{2(k+1)}{x^2+1}\right)\,dx\,dt\\\\ &=-\sum_{n=0}^\infty (-1)^n \log\left(1+\frac1{n+1}\right) \tag2\\\\ &-(2k+2)\int_0^1\int_0^1 \frac{x^t}{(1+x)(1+x^2)}\,dx\\\\ &=-\log(\pi/2)-(2k+2)\int_0^1\int_0^1 \frac{x^t}{(1+x)(1+x^2)}\,dx\tag 3 \end{align}$$

where we used $(1)$ in going from $(2)$ to $(3)$.

Next, using the partial fraction expansion $\frac{1}{(1+x)(1+x^2)}=\frac12\left(\frac{1}{1+x}-\frac{x-1}{x^2+1}\right)$ reveals

$$\begin{align} \int_0^1\int_0^1 \frac{x^t}{(1+x)(1+x^2)}\,dx&=\underbrace{\frac12\int_0^1\int_0^1 \frac{x^t}{1+x}\,dx\,dt}_{=\frac12\log(\pi/2)\,\text{from}\,(1)}-\underbrace{\frac12 \int_0^1\int_0^1 \frac{x^t(x-1)}{1+x^2}\,dx\,dt}_{=-\frac12\log(2)+\frac12 \log(\pi/2)\,\text{from}\,(5)}\tag 4\\\\ &=\frac12\log(2) \end{align}$$

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In arriving at $(4)$, we used the expansion $\frac{1}{1+x^2}=\sum_{n=0}^\infty (-1)^n x^{2n}$. This results in the equality

$$\begin{align} \int_0^1\int_0^1 \frac{x^t(x-1)}{1+x^2}\,dx\,dt&=\sum_{n=0}^\infty (-1)^n \log\left(\frac{2n+3}{2n+2}\frac{2n+1}{2n+2}\right)\\\\ &=\sum_{n=0}^\infty (-1)^n \log\left(\frac{2n+3}{2n+2}\frac{2n+1}{2n+2}\frac{2n+4}{2n+4}\right)\\\\ &=\sum_{n=0}^\infty (-1)^n \log\left(\frac{2n+3}{2n+4}\frac{2n+1}{2n+2}\right)+\sum_{n=0}^\infty (-1)^n\log\left(\frac{n+2}{n+1}\right)\\\\ &=\sum_{n=0}^\infty \left((-1)^n \log\left(\frac{2n+3}{2n+4}\right)-(-1)^{n-1} \log\left(\frac{2(n-1)+3}{2(n-1)+4}\right)\right)+\log(\pi/2)\\\\ &=-\log(2)+\log(\pi/2)\tag 5 \end{align}$$

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Putting it together, we find that

$$\int_0^1 \frac{1-x}{1+x}\frac{x^2+(2k+3)}{x^2+1}\frac{1}{\log(x)}\,dx=-\log(\pi/2)-(k+1)\log(2)=-\log\left(2^{k}\pi\right)$$

Answer 2

By Frullani's theorem $\int_{0}^{+\infty}\frac{e^{-ax}-e^{-bx}}{x}\,dx = \log\frac{b}{a}$ for any $a,b>0$. In particular $$ \int_{0}^{+\infty}\frac{1-e^{-y}}{1+e^{-y}}\cdot\frac{e^{-y}}{y}\,dy $$ equals $\log\frac{\pi}{2}$ since it is the logarithm of Wallis' product. In a similar way $$ \int_{0}^{+\infty}\frac{1-e^{-y}}{1+e^{-y}}\cdot \frac{e^{-y}}{1+e^{-2y}}\cdot\frac{dy}{y} $$ equals $\frac{1}{2}\log(2)$ by partial fraction decomposition, and OP's $(2)$ is straightforward to compute.

Answer 3

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

$\ds{\int_{0}^{1}{1 - x \over 1 + x}\,{2k + 3 + x^{2} \over 1 + x^{2}} \,{\dd x \over \ln\pars{x}} = -\ln\pars{2^{k}\pi}:\ {\large ?}.\qquad k \in \mathbb{R}}$.

\begin{align} &\int_{0}^{1}{1 - x \over 1 + x}\,{2k + 3 + x^{2} \over 1 + x^{2}} \,{\dd x \over \ln\pars{x}} = -\int_{0}^{1}{1 \over 1 + x}\,\pars{1 + 2\,{k + 1 \over 1 + x^{2}}} \,{x - 1 \over \ln\pars{x}}\,\dd x \\[5mm] = &\ -\int_{0}^{1}{1 \over 1 + x}\,{x - 1 \over \ln\pars{x}}\,\dd x - 2\pars{k + 1}\int_{0}^{1}{1 \over 1 + x}\,{1 \over 1 + x^{2}} \,{x - 1 \over \ln\pars{x}}\,\dd x \\[5mm] = &\ -\int_{0}^{1}{1 - x \over 1 - x^{2}}\,{x - 1 \over \ln\pars{x}}\,\dd x - 2\pars{k + 1}\int_{0}^{1}{1 - x \over 1 - x^{4}} \,{x - 1 \over \ln\pars{x}}\,\dd x = \bbx{\ds{a_{1} + 2\pars{k + 1}a_{2}}} \label{1}\tag{1} \\[5mm] &\ \mbox{where}\quad a_{n} \equiv -\int_{0}^{1}{1 - x \over 1 - x^{2n}}\,{x - 1 \over \ln\pars{x}}\,\dd x \label{2}\tag{2} \end{align}

With the identity $\ds{{x - 1 \over \ln\pars{x}} = \int_{0}^{1}x^{t}\,\dd t}$, $\ds{a_{n}}$ becomes:

\begin{align} a_{n} & = \int_{0}^{1}\int_{0}^{1}{x^{t + 1} - x^{t} \over 1 - x^{2n}}\,\dd x\,\dd t \,\,\,\stackrel{x^{2n}\ \mapsto\ x}{=}\,\,\, {1 \over 2n} \int_{0}^{1}\int_{0}^{1} {x^{t/\pars{2n}\ +\ 1/n\ -\ 1}\ -\ x^{t/\pars{2n}\ +\ 1/\pars{2n}\ -\ 1} \over 1 - x}\,\dd x\,\dd t \\[5mm] = &\ {1 \over 2n}\int_{0}^{1}\bracks{% \Psi\pars{t + 1 \over 2n} - \Psi\pars{t + 2 \over 2n}}\,\dd t \end{align} Here, I used Digamma function identity $\ds{\mathbf{6.3.22}}$ in A & S Table. With $\ds{\Psi\pars{z} \equiv \totald{\ln\pars{\Gamma\pars{z}}}{z}}$: \begin{align} a_{n} & = \left.\ln\pars{\Gamma\pars{\bracks{t + 1}/\bracks{2n}} \over \Gamma\pars{\bracks{t + 2}/\bracks{2n}}}\right\vert_{\ t\ =\ 0}^{\ t\ =\ 1} = \ln\pars{{\Gamma\pars{1/n} \over \Gamma\pars{3/\bracks{2n}}}\, {\Gamma\pars{1/n} \over \Gamma\pars{1/\bracks{2n}}}} \\[5mm] = &\ \ln\pars{\Gamma^{2}\pars{1/n} \over \Gamma\pars{3/\bracks{2n}}\Gamma\pars{1/\bracks{2n}}} \\[5mm] \mbox{and} &\ \left\{\begin{array}{rcl} \ds{a_{1}} & \ds{=} & \ds{\ln\pars{\Gamma^{2}\pars{1} \over \Gamma\pars{3/2}\Gamma\pars{1/2}} = -\ln\pars{\pi \over 2}} \\[2mm] \ds{a_{2}} & \ds{=} & \ds{\ln\pars{\Gamma^{2}\pars{1/2} \over \Gamma\pars{3/4}\Gamma\pars{1/4}} = -\ln\pars{\pi \over \pi/\sin\pars{\pi/4}} = -\,{1 \over 2}\,\ln\pars{2}} \end{array}\right.\label{3}\tag{3} \end{align}

With \eqref{1} and \eqref{3}:

\begin{align} &\int_{0}^{1}{1 - x \over 1 + x}\,{2k + 3 + x^{2} \over 1 + x^{2}} \,{\dd x \over \ln\pars{x}} = -\ln\pars{\pi \over 2} + 2\pars{k + 1}\bracks{-\,{1 \over 2}\,\ln\pars{2}} = \bbx{\ds{-\ln\pars{2^{k}\pi}}} \end{align}

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Note that $\ds{\Gamma\pars{1} = 1\,,\ \Gamma\pars{1 \over 2} = \root{\pi}}$ and the use of $\ds{\Gamma}$-Recurrence Property and Euler Reflection Formula.