Integral of the Planck radiation formula

Question

It's well known that the blackbody radiation law has ben derived by Plank and its mathematical formula is: $$W(\lambda,T)=\frac{C_1}{\lambda^5\left(\exp\frac{C_2}{\lambda T}-1\right)}$$ The definite integral of this function: $$J_0(T)=\int_0^\infty W(\lambda,T)d\lambda$$ is known. I have to calculate the following integral: $$J(T)=\int_0^{\lambda_0}W(\lambda,T)d\lambda$$ I tried to make the following change of variables: $\lambda\to \frac{1}{\lambda}=u$. So $d\lambda=-\frac{1}{\lambda^2}du$ and the integral is now $$J(T)=\int_{\frac{1}{u_0}}^\infty W(u,T)du$$ where $u_0=\frac{1}{\lambda_0}$. The result of the indefinite integral $$J_1(T)=\int W(u,T)du$$ is expressed in form of Polylogarithmic and logarithmic functions, so to calculate $J(T)$ I should calculate the limits of these functions. Is this approach correct? Thanks in advance.

Answer 1

Your integral can be cast in the form, after the change of variable you suggested, $$ J(T)=\frac{C_1}{k^4}{\rm Li}_4(1)-\frac{C_1}{k^4}\int_0^{k\lambda_0}\frac{t^3}{e^t-1}dt $$ being ${\rm Li}_n$ the polylogarithm function and I have put $k=\frac{C_2}{T}$. In this form, this function is not difficult to study, even numerically. But Mathematica provides a closed form for the indefinite integral and so $$ J(T)=\frac{C_1}{k^4}\left[\frac{k^4\lambda_0^4}{6}-k^3\lambda_0^3\log\left(1-e^{k\lambda_0}\right)-3k^2\lambda_0^2{\rm Li}_2\left(e^{k\lambda_0}\right)+6k\lambda_0{\rm Li}_3\left(e^{k\lambda_0}\right)-6{\rm Li}_4\left(e^{k\lambda_0}\right)\right]. $$ In this case, one needs to have $e^{k\lambda_0}\le 1$ to keep all real.