Black-body radiation has spectral energy density

Black-body radiation has spectral energy density $E_\lambda=\dfrac{c}{\lambda^5\left(e^{c'/(\lambda T)}-1\right)}$ at absolute temperature $T$, where $c,c'$ are constants. Find the total radiated energy $E=\int_0^\infty E_\lambda\,d\lambda$ as a function of $T$.

Step 1: Substitute $x=\dfrac{c'}{\lambda T}$

Then $\lambda=c'/(xT)$ and $d\lambda=-\dfrac{c'}{x^2T}\,dx$; as $\lambda:0\to\infty$, $x:\infty\to0$. Also $\lambda^5=c'^5/(x^5T^5)$.

Step 2: Rewrite the integral

\[E=\int_0^\infty\frac{c\,d\lambda}{\lambda^5(e^{c'/(\lambda T)}-1)}=\int_0^\infty\frac{c\cdot x^5T^5}{c'^5(e^x-1)}\cdot\frac{c'}{x^2T}\,dx=\frac{c}{c'^4}T^4\int_0^\infty\frac{x^3}{e^x-1}\,dx.\]

(The sign flip from reversing the $x$-limits cancels the sign introduced by $d\lambda$, leaving a manifestly positive integral, as it must be.)

Step 3: Evaluate the dimensionless integral

This is a standard result, expressible via the Gamma and Riemann zeta functions:

\[\int_0^\infty\frac{x^{n-1}}{e^x-1}\,dx=\Gamma(n)\zeta(n)\quad\Longrightarrow\quad \int_0^\infty\frac{x^3}{e^x-1}\,dx=\Gamma(4)\zeta(4)=6\cdot\frac{\pi^4}{90}=\frac{\pi^4}{15}.\]

Step 4: Final result

\[E=\frac{c\pi^4}{15c'^4}\,T^4=\sigma T^4,\qquad \sigma=\frac{c\pi^4}{15c'^4}.\]

\[\boxed{E=\sigma T^4}\]

This is the Stefan–Boltzmann law: total black-body radiated energy grows as the fourth power of absolute temperature. In terms of the physical constants ($c'=hc/k_B$ and absorbing the geometric/angular factors of the full derivation into $c$), this reduces to the familiar $\sigma=\dfrac{2\pi^5k_B^4}{15h^3c^2}\approx5.67\times10^{-8}\ \text{W}\,\text{m}^{-2}\text{K}^{-4}$.

Step 5: Physical consequence

Doubling the absolute temperature multiplies the total radiated energy by $2^4=16$ — the steep temperature dependence that governs stellar luminosity, thermal radiation, and Earth's energy balance.