Evaluation of the integral $\int_0^1 \frac{\ln(1 - x)}{1 + x}dx$

Question

How can I evaluate the integral $$\int_0^1 \frac{\ln(1 - x)}{1 + x}dx$$ I tried manipulating the known integral $$\int_0^1 \frac{\ln(1 - x)}{x}dx = -\frac{\pi^2}{6}$$ but couldn't do anything with it.

Answer 1

You can use double integration:

$$\int\limits_0^1 {\frac{{\log \left( {1 - x} \right)}}{{1 + x}}dx} = \int\limits_0^1 {\int\limits_0^{ - x} {\frac{{du \cdot dx}}{{\left( {1 + u} \right)\left( {1 + x} \right)}}} } $$

$$\int\limits_0^1 {\int\limits_0^x {\frac{{dm \cdot dx}}{{\left( {m - 1} \right)\left( {1 + x} \right)}}} } $$

Now make

$$m = ux $$

$$\int\limits_0^1 {\int\limits_0^1 {\frac{{x \cdot du \cdot dx}}{{\left( {ux - 1} \right)\left( {1 + x} \right)}}} } = \int\limits_0^1 {\int\limits_0^1 {\frac{{du \cdot dx}}{{\left( {ux - 1} \right)}}} } - \int\limits_0^1 {\int\limits_0^1 {\frac{{du \cdot dx}}{{\left( {ux - 1} \right)\left( {1 + x} \right)}}} } $$

We have that (partial fraction decomposition)

$$\frac{1}{ \left( ux - 1 \right)\left( x + 1 \right) } = \frac{u}{ \left( u + 1 \right)\left( ux - 1 \right) } - \frac{1}{ \left( x + 1 \right)\left( u + 1 \right) }$$

So we get

$$\int\limits_0^1 {\int\limits_0^1 {\frac{{du \cdot dx}}{{\left( {ux - 1} \right)}}} } - \int\limits_0^1 {\int\limits_0^1 {\frac{{u \cdot du \cdot dx}}{{\left( {ux - 1} \right)\left( {u + 1} \right)}}} } + \int\limits_0^1 {\int\limits_0^1 {\frac{{du \cdot dx}}{{\left( {x + 1} \right)\left( {u + 1} \right)}}} } $$

Now:

$$\int\limits_0^1 {\int\limits_0^1 {\frac{{du \cdot dx}}{{\left( {ux - 1} \right)}}} } = \int\limits_0^1 {\frac{{\log \left( {1 - u} \right)}}{u}} du = - \frac{{{\pi ^2}}}{6}$$

$$\int\limits_0^1 {\int\limits_0^1 {\frac{{du\cdot dx}}{{\left( {x + 1} \right)\left( {u + 1} \right)}}} } = {\log ^2}2$$

For our last one,note it is the integral we're looking for

$$\int\limits_0^1 {\int\limits_0^1 {\frac{{u\cdot du\cdot dx}}{{\left( {ux - 1} \right)\left( {u + 1} \right)}}} \mathop = \limits^{ux = m} } \int\limits_0^1 {\int\limits_0^u {\frac{{dm\cdot du}}{{\left( {m - 1} \right)\left( {u + 1} \right)}}} } \mathop = \limits^{m = - x} \int\limits_0^1 {\int\limits_0^{ - u} {\frac{{dx\cdot du}}{{\left( {x + 1} \right)\left( {u + 1} \right)}}} } = \int\limits_0^1 {\frac{{\log \left( {1 - u} \right)}}{{ {u + 1} }}} du$$

We get

$$\int\limits_0^1 {\frac{{\log \left( {1 - u} \right)}}{{ {u + 1} }}} du = {\log ^2}2 - \frac{{{\pi ^2}}}{6} - \int\limits_0^1 {\frac{{\log \left( {1 - u} \right)}}{{ {u + 1} }}} du$$

or

$$\int\limits_0^1 {\frac{{\log \left( {1 - u} \right)}}{{{u + 1} }}} du = \frac{{{{\log }^2}2}}{2} - \frac{{{\pi ^2}}}{{12}}$$

as desired.

Answer 2

You can use the integral you want to use, and the Dilogarithm function as mentioned in the comments.

Below we give a complete proof, including a derivation of the value of the integral you wanted to use.

The Dilogarithm function is defined as

$$\text{Li}_2(z) = -\int_{0}^{z} \frac{\log (1-x)}{x} \text{dx} = \sum_{n=1}^{\infty} \frac{z^n}{n^2}, \quad |z| \le 1$$

The integral which you want to use is $\displaystyle -\text{Li}_2(1)$.

Note that $\displaystyle \text{Li}_2(1) = \sum_{n=1}^{\infty} \frac{1}{n^2} = \zeta(2) = \frac{\pi^2}{6}$. (For multiple proofs of that, see here: Different methods to compute $\sum\limits_{k=1}^\infty \frac{1}{k^2}$)

In your integral(whose value you want), make the substitution $\displaystyle x = 2t -1$ and we get

$$\int_{\frac{1}{2}}^{1} \frac{\log (2(1-t))}{t} \text{dt} = \log^2 2 + \int_{\frac{1}{2}}^{1} \frac{\log (1-t)}{t} \text{dt} = \log^2 2 + \text{Li}_2 \left(\frac{1}{2} \right) - \text{Li}_2(1) $$

Now the Dilogarithm function also satisfies the identity

$$\text{Li}_2(x) + \text{Li}_2(1-x) = \frac{\pi^2}{6}-\log x \log (1-x), 0 \lt x \lt 1$$

This identity can easily be proven by just differentiating and using the value of $\displaystyle \text{Li}_2(1)$:

$$\text{Li}_2'(x) - \text{Li}_2'(1-x) = -\frac{\log (1-x)}{x} + \frac{\log x}{1-x} = (-\log x \log (1-x))'$$

and so $$\text{Li}_2(x) + \text{Li}_2(1-x) = C -\log x \log (1-x), 0 \lt x \lt 1$$

Taking limits as $\displaystyle x \to 1$ gives us $\displaystyle C = \frac{\pi^2}{6}$.

Thus

$$\text{Li}_2(x) + \text{Li}_2(1-x) = \frac{\pi^2}{6}-\log x \log (1-x), 0 \lt x \lt 1$$

Setting $\displaystyle x = \frac{1}{2}$ gives us the value of $\displaystyle \text{Li}_2\left(\frac{1}{2}\right) = \frac{\pi^2}{12} - \frac{\log^2 2}{2}$

Thus your integral is

$$\log^2 2 + \text{Li}_2 \left(\frac{1}{2} \right) - \text{Li}_2(1) = \frac{\log^2 2}{2} - \frac{\pi^2}{12}$$

Answer 3

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ $\ds{\int_{0}^{1}{\ln\pars{1 - x} \over 1 + x}\,\dd x = -\,{\pi^{2} \over 6}:\ {\large ?}}$

\begin{align} &\color{#c00000}{\int_{0}^{1}{\ln\pars{1 - x} \over 1 + x}\,\dd x} =\int_{0}^{1}{\ln\pars{x} \over 2 - x}\,\dd x =\int_{0}^{1/2}{\ln\pars{2x} \over 1 - x}\,\dd x \\[3mm]&= \overbrace{\left.\vphantom{\Huge a}-\ln\pars{1 - x}\ln\pars{2x}\right\vert_{0}^{1/2}} ^{\ds{=\ 0}}\ +\ \int_{0}^{1/2}\ln\pars{1 - x}\,{1 \over x}\,\dd x =\color{#c00000}{-\int_{0}^{1/2}{{\rm Li}_{1}\pars{x} \over x}\,\dd x} \end{align} where $\ds{{\rm Li_{s}}\pars{z}}$ is the PolyLogarithm Function. We already used $\ds{{\rm Li_{1}}\pars{z} = -\ln\pars{1 - z}}$.

With the identity ( see the above mentioned link ) $\ds{{\rm Li_{s + 1}}\pars{z} = \int_{0}^{z}{{\rm Li_{s}}\pars{t} \over t}\,\dd t}$ we'll have: $$ \color{#c00000}{\int_{0}^{1}{\ln\pars{1 - x} \over 1 + x}\,\dd x} =\color{#c00000}{-{\rm Li_{2}}\pars{\half}} $$

Also, ( see the above mentioned link ) $\ds{{\rm Li_{2}}\pars{\half} = {\pi^{2} \over 12} - \half\,\ln^{2}\pars{2}}$ which is a consequence of

Euler Reflection Formula $\ds{{\rm Li_{2}}\pars{x} + {\rm Li_{2}}\pars{1 - x} ={\pi^{2} \over 6} -\ln\pars{x}\ln\pars{1 - x}}$.

$$ \color{#00f}{\large\int_{0}^{1}{\ln\pars{1 - x} \over 1 + x}\,\dd x =\half\,\ln^{2}\pars{2} - {\pi^{2} \over 12}} $$

Answer 4

Maple says it's $${(\log2)^2\over2}-{\pi^2\over12}$$ To get there, I think you will have to understand how the known integral you cite was established, and then use the same ideas to do yours (perhaps after first following Emile's calculations).

Answer 5

Following is an elementary proof.

I assume only that $\displaystyle \int_0^1 \frac{\ln x}{1-x}dx=-\frac{\pi^2}{6}$

\begin{align}J&=\int_0^1 \frac{\ln(1-x)}{1+x}dx\\ &\overset{y=\frac{1-x}{1+x}}=\int_0^1 \frac{\ln\left(\frac{2y}{1+y}\right)}{1+y}dy\\ &=\int_0^1 \frac{\ln\left(\frac{2}{1+y}\right)}{1+y}dy+\int_0^1 \frac{\ln t}{1+t}dt\\ &\overset{u=\frac{1-y}{1+y}}=\int_0^1 \frac{\ln\left(1+u\right)}{1+u}du+\int_0^1 \frac{\ln t}{1+t}dt\\ &=\frac{1}{2}\ln^2 2+\int_0^1 \frac{\ln t}{1+t}dt\\ \int_0^1 \frac{\ln t}{1+t}dt&=\int_0^1 \frac{\ln x}{1-x}dx-\int_0^1 \frac{2t\ln t}{1-t^2}dt\\ &\overset{w=t^2}=\int_0^1 \frac{\ln x}{1-x}dx-\frac{1}{2}\int_0^1 \frac{\ln w}{1-w}dw\\ &=\frac{1}{2}\int_0^1 \frac{\ln x}{1-x}dx\\ &=-\frac{1}{12}\pi^2 \end{align} Therefore,

$\boxed{\displaystyle J=\frac{1}{2}\ln^2 2-\frac{1}{12}\pi^2}$

Answer 6

Note: this is not a complete solution, but may serve as a starter

First let $2u=x+1$ and thus $2du=dx$. Then we get: $$\int_0^1\frac{\ln(1-x)}{1+x}dx=\int_{\frac{1}{2}}^1\frac{\ln(2-2u)}{2u}2du$$ $$=\int_{\frac{1}{2}}^1\frac{\ln(2(1-u))}{u}du=\int_{\frac{1}{2}}^1\frac{\ln2+\ln(1-u)}{u}du$$ $$=\int_{\frac{1}{2}}^1\frac{\ln2}{u}du+\int_{\frac{1}{2}}^1\frac{\ln(1-u)}{u}du$$

Answer 7

With subbing $1-x=y$ we have

$$\int_0^1\frac{\ln^a(1-x)}{1+x}dx=\int_0^1\frac{\ln^a(y)}{2-y}dy$$

$$=\sum_{n=1}^\infty\frac{1}{2^n}\int_0^1 y^{n-1}\ln^a(y)dy$$

$$=(-1)^aa!\sum_{n=1}^\infty\frac{1}{2^nn^{a+1}}=(-1)^aa!\text{Li}_{a+1}\left(\frac12\right)$$

Some cases:

By using $\text{Li}_2\left(\frac12\right)=\frac12\zeta(2)-\frac12\ln^2(2)$ and $\text{Li}_3\left(\frac12\right)=\frac78\zeta(3)-\frac12\ln(2)\zeta(2)+\frac16\ln^3(2)$ we have \begin{align} \int_0^1\frac{\ln(1-x)}{1+x}\ dx=-\text{Li}_{2}\left(\frac12\right)=-\frac12\zeta(2)+\frac12\ln^2(2)\label{ln(1-x)/(1+x)} \end{align} \begin{align} \int_0^1\frac{\ln^2(1-x)}{1+x}\ dx=2\text{Li}_{3}\left(\frac12\right)=\frac74\zeta(3)-\ln(2)\zeta(2)+\frac13\ln^3(2)\label{ln^2(1-x)/(1+x)} \end{align} \begin{align} \int_0^1\frac{\ln^3(1-x)}{1+x}\ dx=-6\text{Li}_{4}\left(\frac12\right)\label{ln^3(1-x)/(1+x)} \end{align} \begin{align} \int_0^1\frac{\ln^4(1-x)}{1+x}\ dx=24\text{Li}_{5}\left(\frac12\right)\label{ln^4(1-x)/(1+x)} \end{align}