Homework Help: Deriving the Stefan-Boltzman law and integration tricks

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1. Aug 6, 2015

azoroth134

1. The problem statement, all variables and given/known data
HI people,

I was trying to derive the stefan-boltzman law from the planc's formula, I kind of got stuck with an integral

2. Relevant equations

$$\int_{0}^{\infty} \frac{x^3}{e^x -1} dx$$
3. The attempt at a solution
I tried simplifying it with

$$\int_{0}^{\infty} x^3 e^{-x} \sum_{n=0}^{\infty} e^{-nx} dx$$

Now I don't know what to do with the summation. I could evaluate

$$\int_{0}^{\infty} x^3 e^{-x} dx = 6$$

pleas help me to get the rest of it, I see the answer comes with $\pi$, how do I get it in this kind of equation? Is there any special trick to solve these type of integrals?

2. Aug 6, 2015

ShayanJ

Under the condition that both the series and the integral converge, you can swap them. Here we're sure that the series converges because we got it by taylor-expanding a function. About the integral, we have a polynomial against an exponential with negative exponent which means converges too. So we have $\displaystyle \sum_{n=0}^\infty \int_0^\infty x^3 e^{-x}e^{-nx} dx=\sum_{n=0}^\infty \int_0^\infty x^3 e^{-(n+1)x}$. Now you if you integrate by parts three times(each time considering u to be the algebraic factor and dv to be the exponential one), you can obtain the solution to the integral.

3. Aug 7, 2015

azoroth134

Ah great, thank you very much, I almost got it on the track now, this is

$$\int_{0}^{\infty} x^3 e^{-(n+1)x} dx = 6 (n+1)^{-4}$$

can you just tell me a little about how to evaluate the summation of $$\sum \frac{1}{(n+1)^4}$$ though, I think I forgot this evaluation.

4. Aug 7, 2015

ShayanJ

The series $\displaystyle \sum_{n=0}^\infty \frac{1}{(n+1)^4}$ can be written as $\displaystyle \sum_{n=1}^\infty \frac{1}{n^4}$ which is equal to $\zeta(4)$, where $\zeta(s)$ is the Riemann zeta function. So you don't need to evaluate the series, just look for tables of the values for this function or compute it using math softwares.

5. Aug 8, 2015

azoroth134

problem solved, thanks a lot