Question Number 223241 by Tawa11 last updated on 19/Jul/25
$$\int_{\:\mathrm{0}} ^{\:\mathrm{1}} \:\frac{\mathrm{ln}\left(\mathrm{1}\:−\:\mathrm{x}\right)\:\mathrm{ln}\left(\mathrm{1}\:+\:\mathrm{x}\right)}{\mathrm{x}}\:\mathrm{dx} \\ $$
Answered by altarboy123 last updated on 19/Jul/25
$$=−\frac{\zeta\left(\mathrm{3}\right)}{\mathrm{4}} \\ $$
Commented by Tawa11 last updated on 19/Jul/25
$$\mathrm{No}\:\mathrm{sir}. \\ $$
Commented by Tawa11 last updated on 19/Jul/25
Commented by Tawa11 last updated on 19/Jul/25
$$\mathrm{Wrong}\:\mathrm{sir}. \\ $$
Commented by mnjuly1970 last updated on 20/Jul/25
$$\:\frac{−\mathrm{5}\zeta\left(\mathrm{3}\right)}{\mathrm{8}}=\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} }{{n}^{\mathrm{2}} }\:\mathrm{H}_{{n}} \\ $$
Commented by MathematicalUser2357 last updated on 26/Jul/25
$$\mathrm{but}\:\frac{−\mathrm{5}\zeta\left(\mathrm{3}\right)}{\mathrm{8}}=\mathrm{0}.\mathrm{75128556447474642837483635094465624422811643271281180112016972208864887861644568136653492100583453634137} \\ $$
Commented by Tawa11 last updated on 26/Jul/25
$$\mathrm{you}\:\mathrm{omitted}\:\:\mathrm{minus}\:\:\left(−\:\right) \\ $$
Commented by MathematicalUser2357 last updated on 30/Jul/25
$$− \\ $$
Answered by Nicholas666 last updated on 20/Jul/25
![I = \int_0^1 \frac{\ln(1-x)\ln(1+x)}{x} dx \ln(1-x) = -\sum_{n=1}^{\infty} \frac{x^n}{n} \ln(1+x) = \sum_{m=1}^{\infty} (-1)^{m-1} \frac{x^m}{m} \ln(1-x)\ln(1+x) = \left(-\sum_{n=1}^{\infty} \frac{x^n}{n}\right) \left(\sum_{m=1}^{\infty} (-1)^{m-1} \frac{x^m}{m}\right) = -\sum_{n=1}^{\infty} \sum_{m=1}^{\infty} (-1)^{m-1} \frac{x^{n+m}}{nm} I = \int_0^1 -\sum_{n=1}^{\infty} \sum_{m=1}^{\infty} (-1)^{m-1} \frac{x^{n+m-1}}{nm} dx I = -\sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{(-1)^{m-1}}{nm} \int_0^1 x^{n+m-1} dx \int_0^1 x^{n+m-1} dx = \left[ \frac{x^{n+m}}{n+m} \right]_0^1 = \frac{1}{n+m} I = -\sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{(-1)^{m-1}}{nm(n+m)} \frac{1}{nm(n+m)} = \frac{1}{m^2} \left( \frac{1}{n} - \frac{1}{n+m} \right) I = -\sum_{n=1}^{\infty} \sum_{m=1}^{\infty} (-1)^{m-1} \frac{1}{m^2} \left( \frac{1}{n} - \frac{1}{n+m} \right) I = -\sum_{m=1}^{\infty} \frac{(-1)^{m-1}}{m^2} \sum_{n=1}^{\infty} \left( \frac{1}{n} - \frac{1}{n+m} \right) \sum_{n=1}^{\infty} \left( \frac{1}{n} - \frac{1}{n+m} \right) = \left(1 - \frac{1}{1+m}\right) + \left(\frac{1}{2} - \frac{1}{2+m}\right) + \dots + \left(\frac{1}{m} - \frac{1}{2m}\right) + \dots = H_m I = -\sum_{m=1}^{\infty} \frac{(-1)^{m-1} H_m}{m^2} \sum_{m=1}^{\infty} \frac{(-1)^{m-1} H_m}{m^2} = \frac{5}{8}\zeta(3) I = -\frac{5}{8}\zeta(3) \int_0^1 \frac{\ln(1-x)\ln(1+x)}{x} dx = -\frac{5}{8}\zeta(3)](https://www.tinkutara.com/question/Q223285.png)
$$ \\ $$I = \int_0^1 \frac{\ln(1-x)\ln(1+x)}{x} dx
\ln(1-x) = -\sum_{n=1}^{\infty} \frac{x^n}{n}
\ln(1+x) = \sum_{m=1}^{\infty} (-1)^{m-1} \frac{x^m}{m}
\ln(1-x)\ln(1+x) = \left(-\sum_{n=1}^{\infty} \frac{x^n}{n}\right) \left(\sum_{m=1}^{\infty} (-1)^{m-1} \frac{x^m}{m}\right)
= -\sum_{n=1}^{\infty} \sum_{m=1}^{\infty} (-1)^{m-1} \frac{x^{n+m}}{nm}
I = \int_0^1 -\sum_{n=1}^{\infty} \sum_{m=1}^{\infty} (-1)^{m-1} \frac{x^{n+m-1}}{nm} dx
I = -\sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{(-1)^{m-1}}{nm} \int_0^1 x^{n+m-1} dx
\int_0^1 x^{n+m-1} dx = \left[ \frac{x^{n+m}}{n+m} \right]_0^1 = \frac{1}{n+m}
I = -\sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{(-1)^{m-1}}{nm(n+m)}
\frac{1}{nm(n+m)} = \frac{1}{m^2} \left( \frac{1}{n} – \frac{1}{n+m} \right)
I = -\sum_{n=1}^{\infty} \sum_{m=1}^{\infty} (-1)^{m-1} \frac{1}{m^2} \left( \frac{1}{n} – \frac{1}{n+m} \right)
I = -\sum_{m=1}^{\infty} \frac{(-1)^{m-1}}{m^2} \sum_{n=1}^{\infty} \left( \frac{1}{n} – \frac{1}{n+m} \right)
\sum_{n=1}^{\infty} \left( \frac{1}{n} – \frac{1}{n+m} \right) = \left(1 – \frac{1}{1+m}\right) + \left(\frac{1}{2} – \frac{1}{2+m}\right) + \dots + \left(\frac{1}{m} – \frac{1}{2m}\right) + \dots = H_m
I = -\sum_{m=1}^{\infty} \frac{(-1)^{m-1} H_m}{m^2}
\sum_{m=1}^{\infty} \frac{(-1)^{m-1} H_m}{m^2} = \frac{5}{8}\zeta(3)
I = -\frac{5}{8}\zeta(3)
\int_0^1 \frac{\ln(1-x)\ln(1+x)}{x} dx = -\frac{5}{8}\zeta(3)
\ln(1-x) = -\sum_{n=1}^{\infty} \frac{x^n}{n}
\ln(1+x) = \sum_{m=1}^{\infty} (-1)^{m-1} \frac{x^m}{m}
\ln(1-x)\ln(1+x) = \left(-\sum_{n=1}^{\infty} \frac{x^n}{n}\right) \left(\sum_{m=1}^{\infty} (-1)^{m-1} \frac{x^m}{m}\right)
= -\sum_{n=1}^{\infty} \sum_{m=1}^{\infty} (-1)^{m-1} \frac{x^{n+m}}{nm}
I = \int_0^1 -\sum_{n=1}^{\infty} \sum_{m=1}^{\infty} (-1)^{m-1} \frac{x^{n+m-1}}{nm} dx
I = -\sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{(-1)^{m-1}}{nm} \int_0^1 x^{n+m-1} dx
\int_0^1 x^{n+m-1} dx = \left[ \frac{x^{n+m}}{n+m} \right]_0^1 = \frac{1}{n+m}
I = -\sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{(-1)^{m-1}}{nm(n+m)}
\frac{1}{nm(n+m)} = \frac{1}{m^2} \left( \frac{1}{n} – \frac{1}{n+m} \right)
I = -\sum_{n=1}^{\infty} \sum_{m=1}^{\infty} (-1)^{m-1} \frac{1}{m^2} \left( \frac{1}{n} – \frac{1}{n+m} \right)
I = -\sum_{m=1}^{\infty} \frac{(-1)^{m-1}}{m^2} \sum_{n=1}^{\infty} \left( \frac{1}{n} – \frac{1}{n+m} \right)
\sum_{n=1}^{\infty} \left( \frac{1}{n} – \frac{1}{n+m} \right) = \left(1 – \frac{1}{1+m}\right) + \left(\frac{1}{2} – \frac{1}{2+m}\right) + \dots + \left(\frac{1}{m} – \frac{1}{2m}\right) + \dots = H_m
I = -\sum_{m=1}^{\infty} \frac{(-1)^{m-1} H_m}{m^2}
\sum_{m=1}^{\infty} \frac{(-1)^{m-1} H_m}{m^2} = \frac{5}{8}\zeta(3)
I = -\frac{5}{8}\zeta(3)
\int_0^1 \frac{\ln(1-x)\ln(1+x)}{x} dx = -\frac{5}{8}\zeta(3)