- Thread starter
- #1

- Jan 17, 2013

- 1,667

\(\displaystyle \int^1_0 \frac{\log(t) \log(1-t)}{t} \, dt = \zeta(3)\)

- Thread starter ZaidAlyafey
- Start date

- Thread starter
- #1

- Jan 17, 2013

- 1,667

\(\displaystyle \int^1_0 \frac{\log(t) \log(1-t)}{t} \, dt = \zeta(3)\)

- Thread starter
- #2

- Jan 17, 2013

- 1,667

Here is a small hint

Try integration by parts

- Thread starter
- #3

- Jan 17, 2013

- 1,667

Integrating by parts we get the following

\(\displaystyle \int \frac{\log(x) \log(1-x)}{x}\, dx = -\text{Li}_2(x) \log(x) + \text{Li}_3(x) +C\)

\(\displaystyle \int^1_0 \frac{\log(x) \log(1-x)}{x}\, dx = \text{Li}_3(1) = \zeta(3) \)

- Feb 13, 2012

- 1,704

In...

\(\displaystyle \int^1_0 \frac{\log(t) \log(1-t)}{t} \, dt = \zeta(3)\)

http://www.mathhelpboards.com/f49/integrals-natural-logarithm-5286/

... is reported that if...

$$ f(x) = \sum_{k=0}^{\infty} a_{k}\ x^{k}\ (1)$$

... then...

$$\int_{0}^{1} f(x)\ \ln^{n} x\ d x = (-1)^{n} n!\ \sum_{k=0}^{\infty} \frac{a_{k}}{(k+1)^{n+1}}\ (2)$$

In this case is n=1 and $\displaystyle a_{k}= - \frac{1}{k+1}$, so that...

$$\int_{0}^{1} \frac{\ln (1-x)}{x}\ \ln x\ dx = \zeta (3)\ (3)$$

Kind regards

$\chi$ $\sigma$