Conjecture $\sum_{n=1}^\infty\frac{\ln(n+2)}{n\,(n+1)}\,\stackrel{\color{gray}?}=\,{\large\int}_0^1\frac{x\,(\ln x-1)}{\ln(1-x)}\,dx$

Question

Numerical calculations suggest that $$\sum_{n=1}^\infty\frac{\ln(n+2)}{n\,(n+1)}\,\stackrel{\color{gray}?}=\,\int_0^1\frac{x\,(\ln x-1)}{\ln(1-x)}\,dx=1.553767373413083673460727...$$ How can we prove it? Is there a closed form expression for this value?

Answer 1

By Frullani's theorem $$ \log(n+2)=\int_{0}^{+\infty}\frac{1-e^{-(n+1)x}}{xe^x}\,dx \tag{1}$$ hence by multiplying both sides by $\frac{1}{n(n+1)}$ and summing over $n\geq 1$ we get: $$ \sum_{n\geq 1}\frac{\log(n+2)}{n(n+1)}=\int_{0}^{+\infty}\frac{(1-e^{-x})\left(1-\log(1-e^{-x})\right)}{xe^x}\,dx\tag{2} $$ then, by substituting $x=-\log(u)$: $$ \sum_{n\geq 1}\frac{\log(n+2)}{n(n+1)}=\int_{0}^{1}\frac{(u-1)\left(1-\log(1-u)\right)}{\log(u)}\,du\tag{3} $$ and the claim follows by substituting $v=(1-u)$.

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I do not think there is a nice closed form but in terms of a derivative of some special zeta function, due to $\log(n+2) = \left.\frac{d}{d\alpha}(n+2)^{\alpha}\right|_{\alpha=0^+}$, but for sure $$ \int_{0}^{1}\frac{-x}{\log(1-x)}\,dx = \int_{0}^{1}\frac{x-1}{\log x}\,dx = \log 2$$ and the Taylor series of $\frac{x}{\log(1-x)}$ depends on Gregory coefficients.

Answer 2

Additionnally, by applying Theorem $2$ (here), one obtains a closed form in terms of the poly-Stieltjes constants.

Proposition. We have

$$ \begin{align} \int_0^1\frac{x\,(\ln x-1)}{\ln(1-x)}\,dx= \ln 2+\gamma_1(2,0)-\gamma_1(1,0) \tag1 \\\\\sum_{n=1}^\infty\frac{\ln(n+2)}{n\,(n+1)}=\ln 2+\gamma_1(2,0)-\gamma_1(1,0) \tag2 \end{align}$$

where

$$ \gamma_1(a,b) = \lim_{N\to+\infty}\left(\sum_{n=1}^N \frac{\log (n+a)}{n+b}-\frac{\log^2 \!N}2\right). $$

Answer 3

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Another possible integral representation: \begin{align} \color{#f00}{\sum_{n = 1}^{\infty}{\ln\pars{n + 2} \over n\pars{n + 1}}} & = \sum_{n = 1}^{\infty}{1 \over n\pars{n + 1}}\ \overbrace{\bracks{\pars{n + 1}\int_{0}^{1}{\dd t \over 1 + \pars{n + 1}t}}} ^{\ds{\ln\pars{n + 2}}} \\[3mm] & = \int_{0}^{1} \sum_{n = 0}^{\infty}{1 \over \pars{n + 2 + 1/t}\pars{n + 1}}\,{\dd t \over t} = \int_{0}^{1} {\Psi\pars{2 + 1/t} - \Psi\pars{1} \over 1 + 1/t}\,{\dd t \over t} \\[3mm] & \stackrel{t\ \to 1/t}{=}\ \color{#f00}{\int_{1}^{\infty} {\Psi\pars{2 + t} - \Psi\pars{1} \over t + 1}\,{\dd t \over t} = \gamma\ln\pars{2} + \int_{1}^{\infty} {\Psi\pars{2 + t} \over t\pars{t + 1}}\,\dd t} \end{align}