MHB Stefan-Boltzman law derivation and integral tricks

[clumps tim]

HI people,

I was trying to derive the stefan-boltzman law from the planc's formula, I kind of got stuck with an integral
$$ \int_{0}^{\infty} \frac{x^3}{e^x -1} dx $$

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?
thanks in advance.

 

[Opalg]

HI people,

I was trying to derive the stefan-boltzman law from the planc's formula, I kind of got stuck with an integral
$$ \int_{0}^{\infty} \frac{x^3}{e^x -1} dx $$

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?
thanks in advance.

You are on the right lines. Write the integral as $$\int_{0}^{\infty} x^3 \sum_{n=1}^{\infty} e^{-nx} dx$$. Now take the sum outside the integral (which is justified because the negative exponential makes everything converge rapidly), to get $$ \sum_{n=1}^{\infty}\int_{0}^{\infty} x^3 e^{-nx} dx$$. The integral $$ \int_{0}^{\infty} x^3 e^{-nx} dx$$ can be evaluated by integrating by parts three times, giving the answer $\dfrac6{n^4}.$ Therefore $$\int_{0}^{\infty} \frac{x^3}{e^x -1} dx = \sum_{n=1}^\infty \dfrac6{n^4}.$$ Finally, $$\sum_{n=1}^\infty \dfrac1{n^4}$$ is a well-known series, with sum $\dfrac{\pi^4}{90}.$ So the value of your integral is $\dfrac{\pi^4}{15}$ unless I have made a mistake.

 

[clumps tim]

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.

thanks in advance

 

[Opalg]

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.

Probably the simplest way is to use Parseval's theorem from Fourier theory. See here, for example.