Cubic log integral

[ZaidAlyafey]

Prove the following

\(\displaystyle \int^1_0 \frac{\log^3(1-x)}{x}\, dx=-\frac{\pi^4}{15}\)​

 

[chisigma]

Prove the following

\(\displaystyle \int^1_0 \frac{\log^3(1-x)}{x}\, dx=-\frac{\pi^4}{15}\)​

With the substitution $\displaystyle 1-x=t$ the integral becomes...

$\displaystyle I= \int_{0}^{1} \frac{\ln^{3} t}{1-t}\ dt\ (1)$

... and according to the (5) in...

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

... is...

$\displaystyle I = - 6\ \sum_{k=1}^{\infty} \frac{1}{k^{4}} = - \frac{\pi^{4}}{15}\ (2)$

Kind regards

$\chi$ $\sigma$

 

[ZaidAlyafey]

Actually we can generalize for \(\displaystyle n\geq 2\)

\(\displaystyle

\int^1_0 \frac{\log^{n-1}(1-x)}{x}\, dx =(-1)^{n-1} \int^{\infty}_0 \frac{x^{n-1}}{e^x-1}\,dx = (-1)^{n-1}\Gamma(n) \zeta(n)

\)

For \(\displaystyle n = 4 \)

\(\displaystyle

\int^1_0 \frac{\log^{3}(1-x)}{x}\, dx = -\Gamma(4) \zeta(4)=-6\zeta(4) = -\frac{\pi^4}{15}

\)