MHB Another fun logarithm integral

AI Thread Summary
The discussion presents a mathematical exercise proving that the integral of the product of logarithms, specifically $$\int^1_0 \frac{\log(t) \log(1-t)}{t} \, dt$$ equals the Riemann zeta function at 3, denoted as $$\zeta(3)$$. It suggests using integration by parts to derive the relationship, leading to the conclusion that $$\int^1_0 \frac{\log(x) \log(1-x)}{x}\, dx = \text{Li}_3(1) = \zeta(3)$$. Additionally, it references a formula involving power series and logarithmic integrals, confirming the result with specific values for the series coefficients. The discussion emphasizes the connection between logarithmic integrals and the zeta function, showcasing an interesting mathematical relationship. This integral serves as an engaging exercise for those exploring advanced calculus and number theory.
alyafey22
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Here is a fun exercise to prove

$$\int^1_0 \frac{\log(t) \log(1-t)}{t} \, dt = \zeta(3)$$
 
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Here is a small hint

Try integration by parts
 
We can evaluate the definite integral

Integrating by parts we get the following

$$\int \frac{\log(x) \log(1-x)}{x}\, dx = -\text{Li}_2(x) \log(x) + \text{Li}_3(x) +C$$$$\int^1_0 \frac{\log(x) \log(1-x)}{x}\, dx = \text{Li}_3(1) = \zeta(3) $$
 
ZaidAlyafey said:
Here is a fun exercise to prove

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

In...

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

$\chi$ $\sigma$
 
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