# Integrate $U=\int\limits_{0}^{\infty}\frac{8 \pi hc}{\lambda^5}\frac{1}{e^{hc/\lambda kT}-1} \ d\lambda$

This integral is from a solution of a problem in thermal physics. My prof's notes says the following:

For a particular wavelength, the spectral energydensity is given by Planck's law:

$$u(\lambda)=\frac{8 \pi hc}{\lambda^5}\frac{1}{e^{hc/\lambda kT}-1}\tag1.$$

To find the total energydensity for all wavelengths we have to integrate over all $$\lambda.$$ By letting $$x=hc/\lambda k T$$ we obtain:

$$U=\frac{8\pi k^4T^4}{h^3 c^3}\int\limits_{0}^{\infty}\frac{x^3}{e^x-1} \ dx.\tag2$$

But when I did it, I got the same but a negative value instead.

Doing the same substitution, I get

$$x=\frac{hc}{\lambda k T}\implies dx=-\frac{hc}{kT\lambda^2} \ d\lambda,\tag3$$

so we have

$$\lambda=\frac{hc}{xkT} \quad \text{and} \quad d\lambda=-\frac{kT\lambda^2}{hc} \ dx.\tag4$$

Plugging this in I get

\begin{align}U&=\int\limits_{0}^{\infty}\frac{8 \pi hc}{\lambda^5}\frac{1}{e^{hc/\lambda kT}-1} \ d\lambda=\int\limits_{0}^{\infty}\frac{8 \pi hc}{\lambda^3}\frac{1}{e^{hc/\lambda kT}-1} \ \left(-\frac{kT}{hc}\right)dx \\ &=-\int\limits_{0}^{\infty}\frac{8\pi hc\cdot k^3T^3x^3}{h^3c^3}\cdot\frac{kT}{hc}\frac{1}{e^x-1} \ dx = -\frac{8\pi k^4T^4}{h^3c^3}\int\limits_{0}^{\infty}\frac{x^3}{e^x-1} \ dx. \end{align}

What is wrong with my calculations?

• You didn't change your bounds when you substituted. Commented Oct 23, 2019 at 21:52

Let me do $$I=\int_{0}^{\infty} \frac{x^3 dx}{e^x-1} = \int_{0}^{\infty} \frac{x^3 e^{-x}}{1-e^{-x}} dx =\int_{0}^{\infty} x^3 e^{-x} dx \sum_{k=0}^{\infty} e^{-kx} = \sum_{k=1}^{\infty}\int_{0}^{\infty} x^3 e^{-kx} dx$$ $$\implies I= \sum_{k=1}^{\infty} \frac{3!}{k^4}=6 \zeta(4)= 6 \frac{\pi^4}{90}=\frac{\pi^4}{15}$$