12
$\begingroup$

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

Acutally this is a part of derivation of Stefan-Boltzmann's Law.

And equation should give answer $\frac{\pi^4}{15}$.

$\endgroup$

2 Answers 2

9
$\begingroup$

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, $$6 \sum_{k=1}^\infty k^{-4} = 6 \zeta(4) = \pi^4/15$$

$\endgroup$
7
$\begingroup$

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} $.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.