What is the Black Body Radiation Integral?

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SUMMARY

The discussion centers on evaluating the integral \(\int\frac{x^3}{e^x-1}dx\) from zero to infinity, which is crucial in understanding black body radiation. The integral is transformed into two parts, \(I_1\) and \(I_2\), where \(I_1\) involves the logarithmic function and \(I_2\) results in \(\frac{1}{4}Ln^4(t) + Cte\). The final solution to the integral is established as \(\frac{\pi^4}{15}\), requiring knowledge of geometric series expansion, the Riemann zeta function, and the gamma function. The discussion emphasizes that this problem is non-elementary and references Planck's law for further context.

PREREQUISITES
  • Understanding of integral calculus, particularly improper integrals.
  • Familiarity with the Riemann zeta function and its applications.
  • Knowledge of gamma functions and their properties.
  • Experience with series expansions, especially geometric series.
NEXT STEPS
  • Study the properties and applications of the Riemann zeta function.
  • Learn about gamma functions and their role in advanced calculus.
  • Explore geometric series expansions in mathematical analysis.
  • Review Planck's law and its implications in physics and thermodynamics.
USEFUL FOR

Mathematicians, physicists, and students studying thermodynamics or quantum mechanics who seek to deepen their understanding of black body radiation and related integrals.

project 33.1
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Find\int\frac{x^3}{e^x-1} evaluated between zero and infinitum. I got I=\displaystyle\int\displaystyle\frac{x^3}{e^x-1}dx=\displaystyle\int\displaystyle\frac{Ln^3(t)}{t(t-1)}dt=\displaystyle\int\displaystyle\frac{Ln^3(t)}{t-1}dt-\displaystyle\int\displaystyle\displaystyle\frac{Ln^3(t)}{t}dt=I_1-I_2

I_1=\displaystyle\int\displaystyle\frac{Ln^3(t)}{t-1}dt=\displaystyle\int Ln^3(t)d(Ln(t-1))}

I_2=\displaystyle\int\displaystyle\frac{Ln^3(t)}{t}dt=\displaystyle\frac{1}{4}Ln^4(t)+Cte=\displaystyle\frac{x^4}{4}+Cte

But I can not get near to the solution\pi^4/15
 
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