Integral definida de la ley de Planck

Question

Quiero ser capaz de resolver lo siguiente, \begin{equation} \frac{\int_{0}^{\lambda_0} B(\lambda, T)\,d\lambda} {\int_{0}^{\infty} B(\lambda,T)\,d\lambda} \end{equation}

donde $B(\lambda,T)$ representa la ley de Planck: \begin{equation}\label{planck} B(\lambda,T) = \frac{2 h c^2}{\lambda^5} \frac{1}{\exp{\frac{hc}{\lambda k_B T}} - 1} \end{equation}

La integral del denominador da como resultado la ley de Stefan Boltzmann; sin embargo, ¿cómo podría calcular la integral definida en el numerador?

Answer 1

Con la ayuda del CAS lo he hecho:

$$\int_0^{\text{$ \N - Laambda $0}} \frac{2 h c^2}{\lambda ^5 \left(\exp \left(\frac{h c}{\lambda \text{kB} T}\right)-1\right)} \, d\lambda =\\\int_0^{\text{$ \N - Laambda $0}} \frac{\left(2 h c^2\right) \sum _{j=1}^{\infty } \exp \left(-\frac{j (h c)}{\lambda \text{kB} T}\right)}{\lambda ^5} \, d\lambda =\\\sum _{j=1}^{\infty } \int_0^{\text{$ \N - Laambda $0}} \frac{\left(2 h c^2\right) \exp \left(-\frac{j (h c)}{\lambda \text{kB} T}\right)}{\lambda ^5} \, d\lambda =\sum _{j=1}^{\infty } \frac{2 e^{-\frac{c h j}{\text{kB} T \text{$ \N - Laambda $0}}} \text{kB} T \left(c^3 h^3 j^3+3 c^2 h^2 j^2 \text{kB} T \text{$ \N - Laambda $0}+6 c h j \text{kB}^2 T^2 \text{$ \N - Laambda $0}^2+6 \text{kB}^3 T^3 \text{$ \N - Laambda $0}^3\right)}{c^2 h^3 j^4 \text{$ \N - Laambda $0}^3}=-\frac{2 c \text{kB} T \ln \left(1-e^{-\frac{c h}{\text{kB} T \text{$ \N - Laambda $0}}}\right)}{\text{$ \N - Laambda $0}^3}+\frac{6 \text{kB}^2 T^2 \text{Li}_2\left(e^{-\frac{c h}{\text{kB} T \text{$ \N - Laambda $0}}}\right)}{h \text{$ \N - Laambda $0}^2}+\frac{12 \text{kB}^3 T^3 \text{Li}_3\left(e^{-\frac{c h}{\text{kB} T \text{$ \N - Laambda $0}}}\right)}{c h^2 \text{$ \N - Laambda $0}}+\frac{12 \text{kB}^4 T^4 \text{Li}_4\left(e^{-\frac{c h}{\text{kB} T \text{$ \N - Laambda $0}}}\right)}{c^2 h^3}$$

Código de Mathematica:

Integrate[(2 h c^2)/\[Lambda]^5*1/( Exp[(h c)/(\[Lambda] kB T)] - 1), {\[Lambda], 0, \[Lambda]0}] == -(( 2 c kB T Log[1 - E^(-((c h)/(kB T \[Lambda]0)))])/\[Lambda]0^3) + ( 6 kB^2 T^2 PolyLog[2, E^(-((c h)/(kB T \[Lambda]0)))])/( h \[Lambda]0^2) + ( 12 kB^3 T^3 PolyLog[3, E^(-((c h)/(kB T \[Lambda]0)))])/( c h^2 \[Lambda]0) + ( 12 kB^4 T^4 PolyLog[4, E^(-((c h)/(kB T \[Lambda]0)))])/(c^2 h^3)