# Nice way to find the integral

1. Sep 22, 2008

### roeb

1. The problem statement, all variables and given/known data
$$\int_0^{\inf} \frac{x^3}{e^x - 1}dx$$

2. Relevant equations

3. 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?

2. Sep 22, 2008

### cellotim

Re: Integration

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

3. Sep 22, 2008

### roeb

Re: Integration

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?

4. Sep 22, 2008

### Dick

Re: Integration

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.

Last edited: Sep 22, 2008
5. Sep 23, 2008

### Dick

Re: Integration

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.

Last edited: Sep 23, 2008
6. Sep 23, 2008

### Almanzo

Re: Integration

This is an improper integral, because the integrand x3/(ex-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 ex = u, or x = elog(u). Then x3/(ex-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.

7. Sep 23, 2008

### Dick

Re: Integration

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.

8. Sep 24, 2008

### Almanzo

Re: Integration

Last night, I have thought about it some more.

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

Let Gn(x) = (-1/n)*(x3 + 3x2/n + 6x/n2 + 6/n3)*e-nx
gn(x) = dGn(x)/dx = (x3 + 3x2/n + 6x/n2 +6/n3)*e-nx - (1/n)*(3x2 + 6x/n + 6/n2)*e-nx
gn(x) = x3 * e-nx

Therefore, if dF(x)/dx = x3/(ex-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.

Last edited: Sep 24, 2008
9. Sep 24, 2008

### Dick

Re: Integration

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).

10. Sep 24, 2008

### Dick

Re: Integration

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

11. Sep 24, 2008

### dirk_mec1

Re: Integration

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

12. Sep 25, 2008

### Almanzo

Re: Integration

Oh, yes. Gn(0) = (-1/n)*(6/n3)*e0=(-6/n4). Not zero at all.
But Gn(x) does tend to zero if x increases without bound.
So, the integral of gn(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.)

Last edited: Sep 25, 2008
13. Sep 25, 2008

### cellotim

Re: Integration

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

14. Sep 25, 2008

### Dick

Re: Integration

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).