MHB Prove Cubic Log Integral: -π^4/15

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The integral $$\int^1_0 \frac{\log^3(1-x)}{x}\, dx$$ evaluates to $$-\frac{\pi^4}{15}$$ through a substitution that transforms it into $$I = \int_{0}^{1} \frac{\ln^{3} t}{1-t}\ dt$$. This integral can be computed using the series representation, leading to the result $$I = -6 \sum_{k=1}^{\infty} \frac{1}{k^{4}}$$, which equals $$-\frac{\pi^{4}}{15}$$. Additionally, a generalization for $$n \geq 2$$ shows that $$\int^1_0 \frac{\log^{n-1}(1-x)}{x}\, dx$$ relates to the Gamma function and the Riemann zeta function. Specifically for $$n = 4$$, the integral confirms the earlier result, reinforcing the connection between logarithmic integrals and special functions.
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Prove the following

$$\int^1_0 \frac{\log^3(1-x)}{x}\, dx=-\frac{\pi^4}{15}$$​
 
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ZaidAlyafey said:
Prove the following

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

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

$\chi$ $\sigma$
 
Actually we can generalize for $$n\geq 2$$

$$

\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 $$n = 4 $$

$$

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

$$
 
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