Solving the Integral for Boltzmann Law for Scientists

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Homework Help Overview

The discussion revolves around integrating a specific function related to Planck's formula to derive the Boltzmann law. The integral in question is f(x) = x^3/(e^x-1) evaluated from 0 to infinity, with a known result of pi^4/15.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the integral's relation to the Debye-Einstein integral and mention connections to the polygamma and Riemann zeta functions. There are questions about the elegance of the integration process and the correctness of expressions involving the gamma function and zeta function.

Discussion Status

Some participants have provided insights into the integral's formulation and its connections to known mathematical functions. There is an ongoing exploration of the implications of these connections, with no explicit consensus reached yet.

Contextual Notes

One participant notes their background as an engineer rather than a mathematician, which may influence their approach to the problem. There is also a mention of a typo in a mathematical expression that has been pointed out for clarification.

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:

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

Since:

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

then:

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

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

And thus, would we have:

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

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:

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

(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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