calculus - 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.

Answer

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.