How can I find the integral of \frac{x^3}{e^x - 1} from 0 to infinity?

[roeb]

Homework Statement

$$\int_0^{\inf} \frac{x^3}{e^x - 1}dx$$

Homework Equations

The Attempt at a Solution

I am having a hard time with this one... Not sure what to do

I did the easier integral $$\int \frac{1}{e^x - 1} dx= ln(e^x - 1) - x$$

I then proceeded to attempt integration by parts, but I can't seem to come up with a nice way to find the integral.

$$u = 1/(e^x - 1)....du = \frac{-e^x}{(e^x-1)^2 }dx$$
$$dv = x^3 dx...v = x^4/4$$
But as you can see that seems to complicate the integral so I gave up there...

I also tried by saying:
$$u = x^3...du = 3x^2$$
$$dv = 1/(e^x - 1).....v = ln(e^x - 1) - x$$

But this is even worse...

Integration by parts for this integral doesn't seem to work. I am thinking since the integral goes from 0 to inf that there may be some other trick that I am not aware of. Does anyone know what to do for this integral?

 

[cellotim]

The antiderivative can't be expressed in closed form. Try expanding $$\frac{1}{e^x - 1}$$ into a geometric series.

 

[roeb]

Hmm interesting, thanks. Let me see if I understand correctly,

e^x = 1 + x + ...$$\int_0^{\inf} \frac{x^3}{e^x - 1}dx = \int_0^{\inf} \frac{x^3}{1 + x - 1}dx = \int_0^{\inf} x^2 dx = \inf ?$$

I don't believe I've ever done an integral by using a series substitution like that, but I was under the impression that the answer shouldn't be infinity? Am I doing it incorrectly yet?

 

[Dick]

No, I don't think that's what roeb meant. Use the geometric series expansion 1/(1-e^(-x))=1+e^(-x)+e^(-2x)+... That means 1/(e^x-1)=e^(-x)+e^(-2x)+... This isn't an elementary integral. Now look up the definition of the gamma function. The terms in your expansion look like x^3*exp(-nx). By a change of variable you can put that into the form of the gamma function and get that it's gamma(4)*1/n^4. So you've got gamma(4)*(1/1^4+1/2^4+1/3^4+...). Now look up the definition of the zeta function.

 

[Dick]

The series expansion is $$-\frac{1}{1 - e^x} = -\sum_{n=0}^\infty e^{nx}$$ You can then do integration by parts on each term, $$x^3e^{nx}$$. I have found the answer in several online look-up tables so you can check with those. Be careful taking the alternating series to infinity though. You should get zero.

Zero? That doesn't look right, the integrand is non-negative. Your initial series expansion is divergent and the final series doesn't alternate. Oh, and when I said "I don't think that's what roeb meant" in post 4, I meant "I don't think that's what cellotim meant". The series expansion is good idea. I don't think you've got the parts following that right yet.

 

[Almanzo]

This is an improper integral, because the integrand x³/(e^x-1) is not defined at the edge of the integration domain, x=0. e⁰ - 1 = 1 - 1 = 0. 0³/0 = 0/0 is undefined. So one must first ascertain whether it converges.

To find a primitive function, one might substitute e^x = u, or x = ^elog(u). Then x³/(e^x-1) = {^elog(u)}³/(u-1), and dx = (1/u)du. Integrating by parts, and remembering that 1/{u(u-1)} = 1/(u-1) - 1/u would yield some nice expressions, but also the nasty ^elog(u+1)*d{^elog(u)}. I don't know what to do with that, but perhaps someone else does.

 

[Dick]

This is an improper integral, because the integrand x³/(e^x-1) is not defined at the edge of the integration domain, x=0. e⁰ - 1 = 1 - 1 = 0. 0³/0 = 0/0 is undefined. So one must first ascertain whether it converges.

To find a primitive function, one might substitute e^x = u, or x = ^elog(u). Then x³/(e^x-1) = {^elog(u)}³/(u-1), and dx = (1/u)du. Integrating by parts, and remembering that 1/{u(u-1)} = 1/(u-1) - 1/u would yield some nice expressions, but also the nasty ^elog(u+1)*d{^elog(u)}. I don't know what to do with that, but perhaps someone else does.

Near x=0 it behaves like x^2, roeb's power series expansion shows that clearly enough. x=0 is a removable singularity and nothing to worry about. There is no elementary primitive function. As I've said, you need a zeta function.

 

[Almanzo]

Last night, I have thought about it some more.

x³/(e^x-1) = (x³/e^x)*1/(1-e^-x)
x³/(e^x-1) = (x³/e^x)*(1 + e^-x + e^-2x + e^-3x + e^-4x + ...)
x³/(e^x-1) = x³ * (e^-x + e^-2x + e^-3x + e^-4x + ...)

Let G_n(x) = (-1/n)*(x³ + 3x²/n + 6x/n² + 6/n³)*e^-nx
g_n(x) = dG_n(x)/dx = (x³ + 3x²/n + 6x/n² +6/n³)*e^-nx - (1/n)*(3x² + 6x/n + 6/n²)*e^-nx
g_n(x) = x³ * e^-nx

Therefore, if dF(x)/dx = x³/(e^x-1)
F(x) = G₁(x) + G₂(x) + G₃(x) + G₄(x) + ...

Each of the G's is zero at x=0 and approaches zero if x increases without bound.
But is the same now true for F? This must still be proved.

If it is true for F, then the integral comes to 0 - 0 = 0.

 

[Dick]

Gn(0) isn't 0. It's 6/n^4. (Not -6/n^4). x^3/(exp(x)-1) IS POSITIVE for x>0. How can you even think to claim the integral is zero? It's the sum over n of 6/n^4. It's 6*zeta(4).

 

[Dick]

Ooops. Sorry. You are correct. Gn(0)=(-6/n^4). It's the integral of the nth term that is +6/n^4.

 

[dirk_mec1]

$$\int_{0}^{\infty} \frac{x^3}{e^x-1}\ \mbox{d}x = \int_0^{\infty} \frac{e^{-x}\ x^3}{1-e^{-x}}\ \mbox{d}x = \int_0^{\infty} \sum_{n=0}^{\infty} e^{-(n+1)x}\ x^3\ \mbox{d}x = \int_0^{\infty} \sum_{n=0}^{\infty} \frac{6}{(n+1)^3}\ e^{-(n+1)x}\ \mbox{d}x= \sum_{n=0}^{\infty} \frac{6}{(n+1)^4} = \frac{6 \pi^4}{90}$$

 

[Almanzo]

Oh, yes. G_n(0) = (-1/n)*(6/n³)*e⁰=(-6/n⁴). Not zero at all.
But G_n(x) does tend to zero if x increases without bound.
So, the integral of g_n(x) from 0 to infinity would be (+6/n⁴).

Strangely, because someone had earlier suggested zero as the answer, I accepted a result of zero without carefully checking my own algebra. Sorry.

And, if the integral converges on both ends, the integral of f(x) from 0 to infinity comes to:
6 * (1 + 1/16 + 1/81 + 1/256 + ...). Exactly the sum in the previous post.

(By the way, I wonder how to produce integral and sum signs in my posts.)

 

[cellotim]

Use the tex, /tex commands each in brackets to do latex.

 

[Dick]

Oh, yes. G_n(0) = (-6/n³). Not zero at all.
But G_n(x) does tend to zero if x increases without bound.
So, the integral of g_n(x) from 0 to infinity would be (+6/n³).

Strangely, because someone had earlier suggested zero as the answer, I accepted a result of zero without carefully checking my own algebra. Sorry.

And, if the integral converges on both ends, the integral of f(x) from 0 to infinity comes to:
6 * (1 + 1/8 + 1/27 + 1/64 + ...).

(By the way, I wonder how to produce integral and sum signs in my posts.)

If you click on the symbols in dirk_mec's post, it should pop up a window showing you how the tex symbols were entered. And yes, you've got the integral. The sum of inverse powers is called the zeta function. (As I've said in every post).