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

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.

Solutions Collecting From Web of "Evaluation of the integral $\int_0^1 \frac{\ln(1 – x)}{1 + x}dx$"

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.

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

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

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).

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:
$$\int_0^1\frac{\ln(1-x)}{1+x}dx=\int_{\frac{1}{2}}^1\frac{\ln(2-2u)}{2u}2du$$
$$=\int_{\frac{1}{2}}^1\frac{\ln(2(1-u))}{u}du=\int_{\frac{1}{2}}^1\frac{\ln2+\ln(1-u)}{u}du$$
$$=\int_{\frac{1}{2}}^1\frac{\ln2}{u}du+\int_{\frac{1}{2}}^1\frac{\ln(1-u)}{u}du$$

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$\ds{\int_{0}^{1}{\ln\pars{1 – x} \over 1 + x}\,\dd x = -\,{\pi^{2} \over 6}:\ {\large ?}}$

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

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