Evaluation of the integral: ∫01ln⁡(1−x)1+xdx

Question

Evaluation of the integral:  ∫01ln⁡(1−x)1+xdx

boronganfh · Accepted Answer

You can use double integration:  ∫01log⁡(1−x)1+xdx=∫01∫0−xdu⋅dx(1+u)(1+x)  ∫01∫0xdm⋅dx(m−1)(1+x)  Now make  m=ux  ∫01∫01x⋅du⋅dx(ux−1)(1+x)=∫01∫01du⋅dx(ux−1)−∫01∫01du⋅dx(ux−1)(1+x)  We have that (partial fraction decomposition)  1(ux−1)(x+1)=u(u+1)(ux−1)−1(x+1)(u+1)  So we get  ∫01∫01du⋅dx(ux−1)−∫01∫01u⋅du⋅dx(ux−1)(u+1)+∫01∫01du⋅dx(x+1)(u+1)  Now:  ∫01∫01du⋅dx(ux−1)=∫01log⁡(1−u)udu=−π26

karton · Accepted Answer

Following is an elementary proof. I assume only that  ∫01ln⁡x1−xdx=−π26J=∫01ln⁡(1−x)1+xdx=∫01ln⁡(2y1+y)1+ydy=∫01ln⁡(2y1+y)1+ydy+∫01ln⁡t1+tdt=12ln2⁡2+∫01ln⁡t1+tdt∫01ln⁡t1+tdt=∫01ln⁡x1−xdx−∫012tln⁡t1−t2dt=∫01ln⁡x1−xdx−12∫01ln⁡w1−wdw=12∫01ln⁡x1−xdx=−112π2

user_27qwe · Accepted Answer

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
$L i_{2} (z) = - \int_{0}^{z} \frac{\log (1 - x)}{x} d x = \sum_{n = 1}^{\infty} \frac{z^{n}}{n^{2}}, | z | \leq 1$
The integral which you want to use is $- L i_{2} (1)$
Note that $L i_{2} (1) = \sum_{n = 1}^{\infty} \frac{1}{n^{2}} = ζ (2) = \frac{π^{2}}{6}$ . (For multiple proofs of that, see here: Different methods to compute $\sum_{k = 1}^{\infty} \frac{1}{k^{2}}$ )
In your integral(whose value you want), make the substitution x=2t-1 and we get
$\int_{\frac{1}{2}}^{1} \frac{\log (2 (1 - t))}{t} d t = \log^{2} 2 + \int_{\frac{1}{2}}^{1} \frac{\log (1 - t)}{t} d t = \log^{2} 2 + L i_{2} (\frac{1}{2}) - L i_{2} (1)$
Now the Dilogarithm function also satisfies the identity
$L i_{2} (x) + L i_{2} (1 - x) = \frac{π^{2}}{6} - \log x \log (1 - x), 0 < x < 1$
This identity can easily be proven by just differentiating and using the value of $L i_{2} (1)$ :
$L i_{2}^{'} (x) - L i_{2}^{'} (1 - x) = - \frac{\log (1 - x)}{x} + \frac{\log x}{1 - x}$
$= (- \log x \log (1 - x))^{'}$
and so
$L i_{2} (x) + L i_{2} (1 - x) = С - \log x \log (1 - x), 0 < x < 1$
Taking limits as $x \to 1$ gives us $C = \frac{π^{2}}{6}$
Thus
$L i_{2} (x) + L i_{2} (1 - x) = \frac{π^{2}}{6} - \log x \log (1 - x), 0 < x < 1$
Setting $x = \frac{1}{2}$ gives us the value of $L i_{2} (\frac{1}{2}) = \frac{π^{2}}{12} - \frac{\log^{2} 2}{2}$
Thus your integral is
$\log^{2} 2 + L i_{2} (\frac{1}{2}) - L i_{2} (1) = \frac{\log^{2} 2}{2} - \frac{π^{2}}{12}$

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

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New Questions in Calculus 1