Solving the Integral for Boltzmann Law for Scientists

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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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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]
 
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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)
 
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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