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Cubic log integral

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
Prove the following

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

chisigma

Well-known member
Feb 13, 2012
1,704

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
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}

\)