# A bose-einstein integral

1. May 3, 2014

### ChrisVer

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

I am trying to evuluate the value of the integral:
$J= \int_{0}^{∞} \frac{x^{3}}{e^{x}-1}dx$

Could you please supply me with the method used for that? I thought of breaking the integral from 0 to 1 and from 1 to infinity. That way I could expand the exponential to taylor series for the small values of x, while the second would drop the 1 from the denominator, so it would be like exp[-x]x^3....(am I correct in the idea?).. However with that idea I'm left with a singularity at point 0. Would that mean I need to expand to Laurent series?
(it's not "really" homework, I am just playing around with black body radiation and radiation energy density to show that p[rad]~T^4 which we were given in class).

Last edited: May 3, 2014
2. May 3, 2014

### Staff: Mentor

There should be no singularity at x=0, as the numerator goes to 0 faster than the denominator. I don't see how you get exp[-x]x^3.

3. May 3, 2014

### ChrisVer

Hmm Let me write it in formula:
$\int_{0}^{∞} \frac{x^3}{e^{x}-1} dx= \int_{0}^{1} \frac{x^3}{e^{x}-1} dx+\int_{1}^{∞} \frac{x^3}{e^{x}-1} dx$
so here I'm saying that the first integrand has a singular point at x=0 because the denominator vanishes... Of course you are right about it going slower to 0 for the nominator (I got confused by thinking that exp is a faster function that any power but I lost it hehe). In fact I would expand the exponential to taylor series:
$e^{x} ≈1+x$ and solve to get
$\int_{0}^{1} \frac{x^3}{x}dx= \frac{x^{3}}{3}|^{1}_{0}=1/3$
then I would try to say that the 2nd integrand, for x>1 the denominator is dominated by the exponential rather than the "-1" term... Or in other words I would write:
$\int_{1}^{∞} \frac{x^3}{e^{x}-1} dx =\int_{1}^{∞} \frac{x^3}{e^{x}} dx=\int_{1}^{∞} e^{-x}x^3 dx$

Unfortunately I don't think that this approach can give me the correct result, since the correct result needs π factors... probably numerically I might be close to the correct answer, since I guess it is a game of approximations I use, against a strict result (?)

Last edited: May 3, 2014
4. May 3, 2014

### Ray Vickson

Write
$$f(x) = \frac{x^3}{e^x-1} = x^3 e^{-x} \frac{1}{1 - e^{-x}} = \sum_{n=1}^{\infty} x^3 e^{-nx}$$
Integrate term-by-term. You will get an infinite series with a known sum.

5. May 3, 2014

### Saitama

You do have the right idea about using the taylor series but splitting the integral won't be a very good idea.

$$J=\int_0^{\infty} \frac{x^3}{e^x-1}=\int_0^{\infty} \frac{x^3e^{-x}}{1-e^{-x}}\,dx$$

Now, from the following series expansion,
$$\frac{1}{1-x}=\sum_{k=0}^{\infty} x^k\,\,\,\,\,\, (|x|<1)$$
we can write:
$$\frac{1}{1-e^{-x}}=\sum_{k=0}^{\infty} e^{-kx}$$
Hence,
$$J=\sum_{k=0}^{\infty} \int_0^{\infty} x^3e^{-(k+1)x}\,dx$$

Using integration by parts, you can eliminate $x^3$. I hope this helps.

6. May 3, 2014

### Saitama

I think I found another method which doesn't involve the use of series expansions.

Rewrite:
$$J=\int_0^{\infty} \frac{x^3}{e^x-1}\,dx=\frac{1}{2}\left(\int_0^{\infty} \frac{x^3}{e^{x/2}-1}\,dx-\frac{x^3}{e^{x/2}+1}\,dx\right)=\frac{1}{2}\left(I_1-I_2\right)$$

Proceeding with $I_1$ first, use the substitution $x/2=t$ to get:
$$I_1=\int_0^{\infty} \frac{x^3}{e^{x/2}-1}\,dx=16\int_0^{\infty} \frac{t^3}{e^t-1}\,dt=16J$$
With $I_2$, use the same substitution to get:
$$I_2=\int_0^{\infty} \frac{x^3}{e^{x/2}+1}\,dx=16\int_0^{\infty} \frac{t^3}{e^t+1}\,dt$$
From the formulation of dirichilet eta function,
$$\eta(s)=\frac{1}{\Gamma(s)}\int_0^{\infty} \frac{t^{s-1}}{e^t+1}\,dt$$
$$\Rightarrow \eta(4)=\frac{1}{\Gamma(4)}\int_0^{\infty} \frac{t^3}{e^t+1}\,dt$$
$$\Rightarrow \int_0^{\infty} \frac{t^3}{e^t+1}\,dt =\eta(4)\Gamma(4)$$
Hence,
$$I_2=16\eta(4)\Gamma(4)$$
Therefore,
$$J=\frac{1}{2}\left(16J-16\eta(4)\Gamma(4)\right) \Rightarrow J=\frac{8}{7}\eta(4)\Gamma(4)$$
Since $\eta(4)=7\pi^4/720$ and $\Gamma(4)=3!$, the final result is:
$$\boxed{J=\dfrac{\pi^4}{15}}$$

7. May 3, 2014

### ChrisVer

Pranav, Could you also check Post #3? and give me your "complains"?
One complain I would state for that is the arbitrariness of the function as going from 0 to 1 (I mean exponential would need more powers to be closer to the needed one, like 2nd or 3rd order), the errors at that points ($1^{-}$) would be over 100%....

8. May 3, 2014

### Saitama

I am not sure if I can help with that. I can solve the integrals but I am not very good at doing approximations. I hope someone else can help with that.

9. May 3, 2014

### ChrisVer

nevermind, thanks anyway Shino kun, and the 2nd way was really nice :)

10. May 3, 2014

### Saitama

11. May 5, 2014

### ChrisVer

Well my approximation yields:
$\frac{1}{3}+\frac{16}{e} \approx 6.219401$
$\frac{\pi^{4}}{15} \approx 6.4939394$