Evaluating $\int_{0}^{1}\frac{1-x}{1+x}\frac{\mathrm dx}{\ln x}$

Solution 1:

Make the substitution $x=e^{-y}$ and do a little algebra to get the value of the integral to be

$$\int_0^{\infty} \frac{dy}{y} \frac{e^{-y} - e^{-2 y}}{1+e^{-y}} $$

Now Taylor expand the denominator and get

$$\int_0^{\infty} \frac{dy}{y} (e^{-y} - e^{-2 y}) \sum_{k=0}^{\infty} (-1)^k e^{-k y} $$

If we can reverse the order of sum and integral, we get

$$ \sum_{k=0}^{\infty} (-1)^k \int_0^{\infty} \frac{dy}{y} (e^{-(k+1) y} - e^{-(k+2) y}) $$

The integral inside may be evaluated exactly, and the result is the sum

$$ \sum_{k=0}^{\infty} (-1)^k \log{\frac{k+1}{k+2}} $$

$$ = \lim_{n \rightarrow \infty} \; \log{\frac{\frac{1}{2} \frac{3}{4} \ldots \frac{2 n-1}{2 n}}{\frac{2}{3} \frac{5}{6} \ldots \frac{2 n+1}{2 n+2}}} $$

$$ = \log{\left ( \frac{2}{\pi} \right )} $$

Solution 2:

I tried with "Differentiation under integration sign":

$$J(\alpha)=\int_0^1\frac{1-x^\alpha}{1+x}\frac{dx}{\ln x}\quad \alpha>0$$

Then, as usual, $$\frac{dJ}{d\alpha}=-\int_0^1\frac{x^\alpha}{1+x}dx=-f(\alpha),\text{(say)}$$

Then, integrating $$J(\alpha)=-\int f(\alpha)d\alpha+c$$ I used Mathematica to evaluate $f(\alpha)$ (it gives difference of two Harmonic numbers) and its integral. The integral is just $\ln\frac{\Gamma(\frac{2m+1}{2})}{\Gamma(\frac{m+1}{2})}$. Note that $J(0)=0$ and putting $\alpha=1$, the result follows.

Solution 3:

Consider $$ \mathcal{I}(\alpha)=\int_0^1\frac{1-x^\alpha}{(1+x)\ln x}\ dx.\tag1 $$ Differentiating $(1)$ with respect $\alpha$ yields \begin{align} \frac{d\mathcal{I}}{d\alpha}&=\int_0^1\frac{\partial}{\partial\alpha}\left[\frac{1-x^\alpha}{(1+x)\ln x}\right]\ dx\\ &=-\int_0^1\frac{x^\alpha}{1+x}\ dx\\ &=-\int_0^1\sum_{k=0}^\infty(-1)^kx^{\alpha+k}\ dx\\ &=-\sum_{k=0}^\infty(-1)^k\int_0^1x^{\alpha+k}\ dx\\ &=-\sum_{k=0}^\infty\frac{(-1)^k}{\alpha+k+1}.\tag2 \end{align} Now consider polygamma function $$ \psi_n(z)=\frac{d^{n+1}}{dz^{n+1}}\ln\Gamma(z)\tag3 $$ and $$ \psi_n(z)=(-1)^{n+1}n!\sum_{k=0}^\infty\frac{1}{(z+k)^{n+1}}.\tag4 $$ Hence by using $(4)$ we obtain $$ \sum_{k=0}^\infty\frac{(-1)^{k}}{(z+k)^{n+1}}=\frac1{(-2)^{n+1}n!}\left[\psi_n\left(\frac{z}{2}\right)-\psi_n\left(\frac{z+1}{2}\right)\right].\tag5 $$ Using $(3)$ and $(5)$ then $(2)$ becomes \begin{align} \frac{d\mathcal{I}}{d\alpha}&=\frac12\left[\psi_0\left(\frac{\alpha+1}{2}\right)-\psi_0\left(\frac{\alpha+2}{2}\right)\right]\\ \mathcal{I}(\alpha)&=\frac12\int\left[\psi_0\left(\frac{\alpha+1}{2}\right)-\psi_0\left(\frac{\alpha+2}{2}\right)\right]\ d\alpha\\ &=\ln\Gamma\left(\frac{\alpha+1}{2}\right)-\ln\Gamma\left(\frac{\alpha+2}{2}\right)+C\\ &=\ln\Gamma\left(\frac{\alpha+1}{2}\right)-\ln\Gamma\left(\frac{\alpha+2}{2}\right)-\frac12\ln\pi,\tag6 \end{align} where $\mathcal{I}(0)=0$ and $C=-\ln\Gamma\left(\frac{1}{2}\right)$.

Thus \begin{align} \int_0^1\frac{1-x}{(1+x)\ln x}\ dx&=\mathcal{I}(1)\\ &=\ln\Gamma\left(1\right)-\ln\Gamma\left(\frac{3}{2}\right)-\frac12\ln\pi\\ &=\color{blue}{\ln\left(\frac{2}{\pi}\right)}. \end{align}