How can I calculate $\int_0^\infty \frac{u^3}{(e^u-1)} \, du$?

Since $e^u > 1$, write the integrand as $$ \frac{u^3}{e^u(1 - e^{-u})} = u^3 (e^{-u} + e^{-2u} + e^{-3u} + \ldots)$$ Now for $k > 0$, substituting $t = ku$, $$ \int_0^\infty u^3 e^{-ku} \; du = k^{-4} \int_0^\infty t^3 e^{-t}\; dt = 6 k^{-4}$$ And finally, $$6 \sum_{k=1}^\infty k^{-4} = 6 \zeta(4) = \pi^4/15$$

For $ u \in (0, \infty), 0 < e^{-u} < 1 $ and hence $ \frac{1}{1 - e^{-u}} = \sum\limits_{k = 0}^\infty e^{-ku} $. Therefore: $$ \int_0^\infty \frac{e^{-u} u^3}{1 - e^{-u}}\ du = \int_0^\infty \sum_{k = 1}^\infty u^3 e^{-ku} \ du = \sum_{k = 1}^\infty \int_0^\infty u^3e^{-ku} \ du $$ Next, use repeated integration by parts to arrive at the identity $ \int\limits_0^\infty u^3 e^{-ku} \ du = \frac{6}{k^4} $. A more refined approach than brute force would be to make the substitution $ ku \mapsto x $ and use the Gamma function, specifically $ \int\limits_0^\infty x^3 e^{-x} \ dx = 3! = 6 $.

Hence, the sum is simply $ \sum\limits_{k = 1}^\infty \frac{6}{k^4} = 6 \zeta(4) = \frac{\pi^4}{15} $.