Integral equation involving Planck radiation formula

Question

I am stuck in solving the following integral equation: $$\sigma T^4=\pi\int_{\lambda_0}^{\lambda_1}d\lambda W_{\lambda,T}$$ where: $$W_{\lambda,T}=\dfrac{C_1}{\lambda^5\left(\exp\left(\frac{C_2}{\lambda T}\right)-1\right)}$$ and $C_1,C_2$ are constant coefficients. Fixing $\lambda_1$ and $\lambda_2$ and plotting the left and the right side of the equation, I found the approximate numerical value of the variable $T$. Is it possible to solve the previous equation analytically? Thanks.

Answer 1

Ok as suggested in my comment, transform $b/\lambda=x, d\lambda=-b/x^2$ with $b=C_2/T$. Therefore our integral becomes:

$$ I(C_1,b,\lambda_1,\lambda_0)=\frac{C_1 \pi}{b^4}\underbrace{\int^{b\lambda_1}_{b\lambda_0}\frac{x^3}{1-e^x}}_{J(b\lambda_1,b\lambda_0)}dx $$

therefore (i will not care about possible convergence issues, i leave this to you but there is no real difficulty) using the geometric series $$ {J(b\lambda_1,b\lambda_0)}=\sum_{n=0}^{\infty}\int^{b\lambda_1}_{b\lambda_0}x^3e^{nx}dx=\left[\frac{x^4}{4}-\sum_{n=1}^{\infty}\frac{6e^{nx}}{n^4}-\frac{6xe^{nx}}{n^3}+\frac{3x^2e^{nx}}{n^2}-\frac{x^3 e^{nx}}{n}\right]_{x=\lambda_0b}^{{x=\lambda_1b}} $$

You may conclude by using the Definition of Polylogarithm

$$Li_s(z)=\sum_{n=1}^{\infty}\frac{z^n}{n^s}$$

Because the result is rather messy, i spare it here.