How to prove $\int_0^1\frac{1-x}{(\ln x)(1+x)}\ dx=\ln\left(\frac2{\pi}\right)$?

Question

A friend asked me to prove

$$\int_0^1\frac{1-x}{(\ln x)(1+x)}\ dx=\ln\left(\frac2{\pi}\right)$$

and here is my approach:

\begin{align} I&=\int_0^1\frac{1-x}{\ln x}\frac1{1+x}\ dx\\ &=\int_0^1\left(-\int_0^1x^y\ dy\right)\frac1{1+x}\ dx\\ &=\int_0^1\left(-\int_0^1\frac{x^y}{1+x}\ dx\right)\ dy\\ &=\int_0^1\left((-1)^n\sum_{n=1}^\infty\int_0^1x^{y+n-1}\ dx\right)\ dy\\ &=\int_0^1\left(\sum_{n=1}^\infty\frac{(-1)^n}{y+n}\right)\ dy\\ &=\sum_{n=1}^\infty(-1)^n\int_0^1\frac1{y+n}\ dy\\ &=\sum_{n=1}^\infty(-1)^n\left[\ln(n+1)-\ln(n)\right]\tag{1} \end{align}

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Now how can we finish this alternating sum into $\ln\left(\frac2{\pi}\right)$?

My idea was to use

$$\operatorname{Li}_a(-1)=(1-2^{-a})\zeta(a)=\sum_{n=1}^\infty\frac{(-1)^n}{n^a}$$

and if we differentiate both sides with respect to $a$ we get

$$2^{1-a}(\zeta^{'}(a)-\ln2\zeta(a))-\zeta^{'}(a)=\sum_{n=1}^\infty\frac{(-1)^{n-1}\ln(n)}{n^a}\tag{2}$$

wolfram says that $\sum_{n=1}^\infty(-1)^n\ln(n)$ is divergent which means we can not take the limit for (2) where $a\mapsto 0$ which means we can not break the sum in (1) into two sums. So any idea how to do (1)?

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Other question is, I tried to simplify the sum in (1) as follows

\begin{align} S&=\sum_{n=1}^\infty(-1)^n\left[\ln(n+1)-\ln(n)\right]\\ &=-\left[\ln(2)-\ln(1)\right]+\left[\ln(3)-\ln(2)\right]-\left[\ln(4)-\ln(3)\right]+\left[\ln(5)-\ln(4)\right]-...\\ &=-2\ln(2)+2\ln(3)-2\ln(4)+...\\ &=2\sum_{n=1}^\infty(-1)^n\ln(n+1) \end{align}

which is divergent again. what went wrong in my steps?

Thanks.

Answer 1

We've $\displaystyle \sum_{n \ge 1} (-1)^n \left[\log(n+1)- \log(n)\right] = \sum_{n \ge 1}(-1)^n\log\left(\frac{n+1}{n}\right) = \log(\mathcal{P})$ where:

$\displaystyle \mathcal{P} = \prod_{n \ge 1}\bigg(\frac{n+1}{n}\bigg)^{(-1)^n} = \prod_{n \ge 1} \bigg(\frac{2n+1}{2n}\bigg) \bigg(\frac{2n}{2n-1}\bigg)^{-1} = \prod_{n \ge 1} \bigg(\frac{2n+1}{2n}\cdot \frac{2n-1}{2n}\bigg) = \frac{2}{\pi}. $

Where the last step is the Wallis product for $\pi$.

Answer 2

Define $I(a) := \int_0^1 \frac{x^a(1-x)}{\ln(x)(1+x)}dx$. Then, for $a > -1$, $I'(a) = \int_0^1 \frac{x^a(1-x)}{1+x}dx = \frac{2_2F_1(1,1+a;2+a;-1)-1}{1+a}$, so $I(a) = 2\log \left(\Gamma(\frac{a}{2}+1)\right)-2\log\left(\Gamma(\frac{a+1}{2})\right)-\log(a+1)+\log(2)$. Plug in $a=0$.

Answer 3

Note that $$\int_0^1 x^y dy=\frac{x-1}{\ln x}$$using this result our integral becomes $$\begin{aligned} I &=\int_0^1\frac{x-1}{(x+1) \ln x}\\&=\int_0^1\left(\frac{1}{x+1}\int_0^1x^ydy\right)dx \\&=\int_0^1\int_0^1\frac{x^y}{1+x}dy dx\\&=\int_0^1\underbrace{\int_0^1\frac{x^y}{1+x}dx }_{I_1}dy \end{aligned}$$ Since the integral $$I_1=\int_0^1\frac{x^y}{1+x}dx =\frac{1}{2}\left(H_{\frac{ y}{2}}-H_{\frac{y-1}{2}}\right)=\frac{1}{2}\left(\psi^0\left(\frac{2y+1}{2}\right)-\psi^0\left(\frac{y+1}{2}\right)\right)$$ is well know result which I have proved here by polynomial long division. Integrating $I_1$ we yield $$ \begin{aligned}I &= \int_0^1 I_1 dy \\&=\left[\ln\Gamma\left(\frac{y+2}{2}\right)-\ln\Gamma\left(\frac{y+1}{2}\right)\right]_0^1\\& =\ln\left[ \Gamma\left(\frac{3}{2}\right)\Gamma\left(\frac{1}{2}\right)\right)\cdots(1)\\&=\ln\left(\frac{\sqrt{\pi}}{2}\sqrt{\pi}\right)=\ln\left(\frac{\pi}{2}\right)\end{aligned}$$ we use the half gamma $\displaystyle \Gamma\left(\frac{1}{2}+n\right)=\frac{(2n)!}{4^n n!}\sqrt{\pi}$ argument in $(1)$ or using functional equation of gamma function we can write $(1)$ as $\displaystyle=\ln\left(\frac{1}{2}\Gamma^2\left(\frac{1}{2}\right)\right)=\ln\left(\frac{\pi}{2}\right)$

Alternatively$$\begin{aligned}I_1 & =\int_0^1\frac{x^y}{1+x}dx \\&=\int_0^1x^y \left(\sum_{r=0}^{\infty} (-1)^r x^r\right)dx\\&=\sum_{r=0}^{\infty} \frac{(-1)^r}{y+r+1}=\Phi\left(-1,1,y+1\right)\cdots(3)\end{aligned}$$ where $\Phi(z,s,\alpha )$ is Lerch Transcendent function using the equation 5 and 6 we obtain $$I= \frac{1}{2}\left(\psi^0\left(\frac{2y+1}{2}\right)-\psi^0\left(\frac{y+1}{2}\right)\right)$$ To prove tha relationship between $ (3)$ and final result we can directly use series formula of digamma function.

The above integral can be related as

$$\int_0^{\infty} \frac{\operatorname{tanh}x}{e^{2x}} \frac{dx}{x}=\int_0^1\frac{x-1}{(1+x)\ln x} =-\ln\left(\frac{2}{\pi}\right)$$

Answer 4

Let

$$I=\int_0^1\frac{1-x}{\ln x(1+x)}\ dx$$

and $$I(a)=\int_0^1\frac{1-x^a}{\ln x(1+x)}\ dx.$$

Note that $I(1)=I$ and $I(0)=0.$

$$I'(a)=\int_0^1\frac{-x^a}{1+x}\ dx=\frac12\psi\left(\frac{a+1}{2}\right)-\frac12\psi\left(\frac{a+2}{2}\right)$$

$$\Longrightarrow \int_0^1I'(a)da=I(1)-I(0)=I=\frac12\int_0^1 \left[\psi\left(\frac{a+1}{2}\right)-\psi\left(\frac{a+2}{2}\right)\right]da $$

$$=\left[\ln\Gamma\left(\frac{a+1}{2}\right)-\ln\Gamma\left(\frac{a+2}{2}\right)\right]_0^1=\ln\frac{\Gamma(\frac{a+1}{2})}{\Gamma(\frac{a+2}{2})}\Bigg|_0^1$$

$$=-\ln\Gamma\left(\frac32\right)-\ln\Gamma\left(\frac12\right)=-\ln \left(\Gamma\left(\frac32\right)\Gamma\left(\frac12\right)\right)$$

$$=-\ln \left(\frac12\Gamma^2\left(\frac12\right)\right)=-\ln\left(\frac{\pi}{2}\right).$$