What is the result of $\int_{0}^{\infty} \frac{1}{x(x+1)}\ln(x+1)dx$

Question

Is there a result for $$ \int_{0}^{\infty} \frac{1}{x(x+1)}\ln(x+1)dx $$ If not, is there any upper bound for that?

Update: The result is $\frac{\pi^2}{6}$. But how to prove it?

Answer 1

Use substitution $u=\frac{x}{x+1}$. Then $x=1-\frac{1}{1-u}$ and \begin{align} \int_{0}^{\infty}\frac{1}{x(x+1)}\ln(x+1)dx=\int_{0}^{1}\frac{1}{u}\ln\left(\frac{1}{1-u}\right)du=Li_{2}(1)=\frac{\pi^{2}}{6} \end{align} where $Li_{2}(z)$ is dilogarithm.

Answer 2

Let $u = \ln(x+1) \implies x = e^u-1,dx = e^udu, x+1 = e^u $

The integral becomes: $$\int_{0}^{\infty} \frac{1}{x(x+1)}\ln(x+1)dx= \begin{equation} \int_0^{\infty} \frac{u}{e^u-1} du \end{equation}$$

Which is similar to this post which shows that $\begin{equation} \int_0^{\infty} \frac{u}{e^u-1} du \end{equation} = \frac{\pi^2}{6}$

Answer 3

Let $I$ denote the integral. Substituting $u = \ln(x+1)$, we get:

$$I = \int_0^{\infty} \frac{u}{e^u -1 }du$$

So:

$$I = -\int_0^{\infty} \frac{ue^{-u}}{1- e^{-u}} du = - \int_0^{\infty} ue^{-u} \sum_{n=0}^{\infty} (e^{-u})^n du = -\sum_{n=0}^{\infty} \int_0^{\infty} ue^{-(n+1)u}du$$

Integrating by parts,

$$\int_0^{\infty} ue^{-(n+1)u}du = -\frac1{(n+1)^2}$$

Hence, $I = \sum_{n=0}^{\infty} \frac1{(n+1)^2} = \pi^2/6$

Answer 4

Let the integral be denoted as $I$. Make the substitution $u={x+1}$ and $du=dx$

we have $$I=\int_{1}^{\infty}\frac{\ln(u)}{(u-1)u}du.$$ Now let $u=\frac{1}{z}$ so that $du=\frac{-1}{z^2}dz.$

Now $I=\int_{0}^{1} \frac{\ln(z)}{z-1}dz.$

Now rewrite the integrand as a geometric series as such $$\frac{\ln(z)}{z-1}=-\sum_{n=0}^{\infty}\ln(z)z^n.$$

Replace this with the integrand in $I.$ Now we will integrate term by term.

Use the identity $\int_{0}^{1} \ln(z)z^n dz=-\frac{1}{(n+1)^2},$ which you can derive by integration by parts or differentiation under the integral sign.

Thus, we have $$I=\sum_{n=0}^{\infty} \frac{1}{(n+1)^2}=\sum_{n=1}^{\infty} \frac{1}{n^2}.$$

To find the exact value of $I$, which is $\frac{\pi^2}{6}$ look at this http://math.cmu.edu/~bwsulliv/MathGradTalkZeta2.pdf

Answer 5

$$\begin{align}\int_0^\infty\frac{\ln(1+x)}{x(1+x)}\mathrm{d}x &= \int_0^1\frac{\ln(1+x)}{x(1+x)}\mathrm{d}x + \int_1^\infty\frac{\ln(1+x)}{x(1+x)}\mathrm{d}x \\ &= \int_0^1\frac{\ln(1+x)}{x(1+x)}\mathrm{d}x + \int_0^1\frac{\ln(1+\frac{1}{x})}{\frac{1}{x}(1+\frac{1}{x})x^2}\mathrm{d}x\tag{1} \\ &= \int_0^1\frac{\ln(1+x)}{x}-\frac{\ln(1+x)}{1+x}\mathrm{d}x + \int_0^1\frac{\ln(1+x)-\ln x}{1+x}\mathrm{d}x\tag{2}\\ &= -\sum_{k=1}^\infty\int_0^1\frac{(-1)^k x^k}{kx}\mathrm{d}x- \sum_{k=0}^\infty\int_0^1(-1)^k x^k\ln x\mathrm{d}x\tag{3}\\ &= -\sum_{k=1}^\infty\frac{(-1)^k}{k^2}-\sum_{k=0}^\infty\frac{\mathrm{d}}{\mathrm{d}a}\int_0^1(-1)^k x^a\mathrm{d}x\bigg{|}_{a=k}\tag{4}\\ &= -\sum_{k=1}^\infty\frac{(-1)^k}{k^2}-\sum_{k=0}^\infty\frac{\mathrm{d}}{\mathrm{d}a}\frac{(-1)^k}{a+1}\bigg{|}_{a=k}\\ &= -\sum_{k=1}^\infty\frac{(-1)^k}{k^2}+\sum_{k=0}^\infty\frac{(-1)^k}{(k+1)^2}\\ &= -\sum_{k=1}^\infty\frac{(-1)^k}{k^2}-\sum_{k=1}^\infty\frac{(-1)^k}{k^2}\tag{5} \\ &= 2\cdot\frac{\pi^2}{12}=\frac{\pi^2}{6}\tag{6} \end{align}$$

Explanations :

$(1)$ Substitute $x\rightarrow 1/x$

$(2)$ Use $1/(x(1+x))=1/x-1/(1+x)$

$(3)$ Expand $ln(1+x)$ and $1/(1+x)$ in Taylor series

$(4)$ Use $x^k\ln x = \frac{\mathrm{d}}{\mathrm{d}a}x^a\bigg{|}_{a=k}$

$(5)$ Shift $k+1\rightarrow 1$

$(6)$ use $\sum_{k=1}^\infty (-1)^k/k^2 = -\pi^2/12$

Answer 6

$\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,\mathrm{Li}_{#1}} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \color{#f00}{\int_{0}^{\infty}{\ln\pars{x + 1} \over x\pars{x + 1}}\,\dd x} &\ \stackrel{\pars{x + 1}\ \mapsto\ x}{=}\ \int_{1}^{\infty}{\ln\pars{x} \over \pars{x - 1}x}\,\dd x\ \stackrel{x\ \mapsto\ 1/x}{=}\,\,\,\, -\int_{0}^{1}{\ln\pars{x} \over 1 - x}\,\dd x \\[5mm] &\ \stackrel{x\ \mapsto\ \pars{1 - x}}{=}\,\,\,\,\, \underbrace{-\int_{0}^{1}{\ln\pars{1 - x} \over x}\,\dd x} _{\ds{\color{#f00}{\large ?}}}\ =\ \int_{0}^{1}\Li{2}'\pars{x}\,\dd x = \Li{2}\pars{1} = \color{#f00}{\pi^{2} \over 6} \end{align}

Otherwise, $$ \color{#f00}{\large ?} = -\int_{0}^{1}{\ln\pars{1 - x} \over x}\,\dd x = \sum_{n = 1}^{\infty}{1 \over n}\int_{0}^{1}x^{n - 1}\,\dd x = \sum_{n = 1}^{\infty}{1 \over n^{2}} = \color{#f00}{\pi^{2} \over 6} $$