# How to solve this integral

1. Dec 2, 2005

### jaap de vries

When integrating plank's fomula to obatin boltzman 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

2. Dec 2, 2005

### George Jones

Staff Emeritus
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

3. Dec 2, 2005

### saltydog

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: Dec 2, 2005
4. Dec 2, 2005

### George Jones

Staff Emeritus
Very nice.

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

5. Dec 2, 2005

### saltydog

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: Dec 2, 2005
6. Dec 2, 2005

### jaap de vries

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