Solving the Integral for Boltzmann Law for Scientists

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SUMMARY

The integral of the function f(x) = x^3/(e^x-1) from 0 to infinity results in pi^4/15, a key aspect of deriving the Boltzmann law. This integral is identified as the Debye-Einstein integral, denoted as \mathcal{D}_3. The solution involves the use of the gamma function and the Riemann zeta function, specifically \Gamma(4) and \zeta(4). The discussion emphasizes the relationship between the integral and the formula \int_0^{\infty}\frac{x^n}{e^x-1}dx = \Gamma(n+1)\zeta(n+1>.

PREREQUISITES
  • Understanding of integral calculus, particularly improper integrals.
  • Familiarity with the gamma function and its properties.
  • Knowledge of the Riemann zeta function and its applications.
  • Basic concepts of statistical mechanics related to the Boltzmann law.
NEXT STEPS
  • Study the properties and applications of the gamma function in integration.
  • Learn about the Riemann zeta function and its significance in number theory and physics.
  • Explore the derivation and applications of the Debye-Einstein integral.
  • Investigate the relationship between statistical mechanics and thermodynamic integrals.
USEFUL FOR

Scientists, mathematicians, and engineers interested in statistical mechanics, particularly those working on problems related to the Boltzmann law and integrals in theoretical physics.

jaap de vries
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When integrating plank's fomula to obatin Boltzmann law,
I need to integrate

f(x) = x^3/(e^x-1) from 0 to infinity, the result is pi^4/15

Does anybody have any idea on how to do this elegantly??
Thank you.
Jaap
 
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This is a problem in Arfken. It involves the ploygamma function and the Riemann zeta function. I won't have time to think about it until later today.

Regards,
George
 
jaap de vries said:
When integrating plank's fomula to obatin Boltzmann law,
I need to integrate
f(x) = x^3/(e^x-1) from 0 to infinity, the result is pi^4/15
Does anybody have any idea on how to do this elegantly??
Thank you.
Jaap

This is a problem previously addressed by Daniel:

The integral is the Debye-Einstein integral:

\mathcal{D}_3=\int_0^{\infty} \frac{x^3}{e^x-1}dx=\int_0^\infty \frac{x^3e^{-x}}{1-e^{-x}}dx

Since:

\frac{1}{1-e^{-x}}=\sum_{n=0}^{\infty} \left(\frac{1}{e^x}\right)^n

then:

\sum_{n=1}^{\infty}\int_0^{\infty} x^3 e^{-nx}dx=\Gamma(x)\zeta(4)
 
Last edited:
saltydog said:
This is a problem previously addressed by Daniel

Very nice. :smile:

Picking a nit - there's a minor typo in the last line.

I had hoped to have a go at this problem this afternoon after finishing my "real" work; now I guess I'll have to find something else to do.

Regards,
George
 
George Jones said:
Very nice. :smile:
Picking a nit - there's a minor typo in the last line.
Regards,
George

Thanks for pointing that out. Should it read:

\sum_{n=1}^{\infty}\int_0^{\infty} x^3 e^{-nx}dx=\Gamma(4)\zeta(4)

And thus, would we have:

\mathcal{D}_n=\Gamma(n+1)\zeta(n+1)\quad ?

I'm not sure and will need to look at it a bit. Well, . . . how about you Jaap?

Edit:

Yep, yep, I think we should re-phrase the question:

Japp, kindly prove or disprove the following:

\int_0^{\infty}\frac{x^n}{e^x-1}dx \:?=\:\Gamma(n+1)\zeta(n+1)

(and he also showed me how to put that question mark on top of the equal sign but I forgot)
 
Last edited:
Thanks guys! let me chew on that one a bit. Note however, I'm an engineer not a mathematician. Nice to know there is a community out here to help, Makes me feel good.

I'll let Y'all know if I have any questions

Jaap
 

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