# Integral problem

1. Jun 25, 2005

### vabamyyr

im havin trouble evaluating integral $$\int_0^\infty\frac{x^{3}}{e^x-1}dx$$

i tried integrating by substitution but got nowhere. if there wasnt factor -1 it would equal G(4) which is 3! but first integral equals pii^4/15 which is 6,49....

But any hint how to do this integral?

Last edited: Jun 25, 2005
2. Jun 25, 2005

### dextercioby

Expand $$\frac{1}{e^{x}-1}$$ into series and part integrate.

Daniel.

3. Jun 25, 2005

### Jameson

You could expand it to a series or you could integrate that and then take the limit. This will require integration by parts, but the tabular method will make it less messy with trying to work out the different "u's" and "dv's".

$$\int_0^\infty\frac{x^{3}}{e^x-1}dx = \lim_{b\rightarrow\infty}\int_0^{b}\frac{x^{3}}{e^x-1}dx$$

Jameson

4. Jun 26, 2005

### vabamyyr

$$\int_0^\infty\frac{x^{3}}{e^x-1}dx = \lim_{b\rightarrow\infty}\int_0^{b}\frac{x^{3}}{e^{x}-1}dx=6(lix - xlix+0.5x^{2}lix+\frac{1}{6}x^{3}ln(1-e^{x})-\frac{1}{4}x^{4}+C$$ from 0 to b $$lix=\int\frac{dx}{lnx}$$ from 0 to infinity but i have trouble evaluating that too

i tried ur idea aswell dextercioby by trying to expand $$\frac{1}{e^{x}-1}$$ i figured out that $${e^{x}-1}=\Sigma_{1}^{\infty}\frac{x^k}{k!}$$
but got stuck with $$\frac{1}{e^{x}-1}$$

Last edited: Jun 26, 2005
5. Jun 26, 2005

### dextercioby

I told you how to do it.This is the famous Debye-Einstein integral.

$$\mathcal{D}_{3}=\int_{0}^{\infty} \frac{x^{3}}{e^{x}-1} \ dx=\int_{0}^{+\infty} \frac{x^{3}e^{-x}}{1-e^{-x}} \ dx=...=\sum_{n=1}^{\infty} \int_{0}^{\infty} x^{3}e^{-nx} \ dx =\Gamma (4)\zeta (4)$$

Daniel.

6. Jun 26, 2005

### vabamyyr

7. Jun 26, 2005

### dextercioby

$$\frac{1}{1-e^{-x}} =\sum_{n=0}^{\infty} e^{-nx}$$

is all you need.

Daniel.

8. Jun 27, 2005

### vabamyyr

thx dextercioby, i learned little about Riemann zeta function and now i see where pi^4/15 comes from

9. Jun 27, 2005

### saltydog

Thanks Daniel. Interesting problem and approach. Took me a while to figure it out:

$$\sum_{n=1}^{\infty} \int_{0}^{\infty} x^{3}e^{-nx} \ dx =\sum_{n=1}^{\infty}\frac{6}{n^4}$$

For the sum, Mathematica returns:

$$=\frac{\pi^4}{15}$$

As far as:

$$\sum_{n=1}^{\infty}\frac{6}{n^4}=\Gamma (4)\zeta (4)$$

Suppose I can dig into those too.

Edit: Alright, I don't know why I got the Zeta function mixed up with a sum of primes.
Nevermind.

Edit2: Here embrass me:

$$\zeta(n)=\sum_{k=1}^{\infty}\frac{1}{k^n}$$

IS the Zeta function dude!

And:

$$\Gamma(4)=6$$

Last edited: Jun 27, 2005
10. Jun 27, 2005

### vabamyyr

there are several shapes to express zeta function and u dont have to tell me what gamma function is

11. Jun 27, 2005

### shmoe

It can be expresssed as a certain product over primes, you could have been thinking of that.

One comment I'd like to make is about the ... in Daniel's post. This is leaving out the important technical bit of justifying switching the infinite sum with the integral (which isn't always valid, but is here of course). I'm sure he's aware of this, but anyone trying to fill in the ... should keep this in mind.

12. Jun 27, 2005

### saltydog

Vabamyyr, sorry you mis-understood my comments: I was saying that to me. :yuck: Not you. You knew it before me.

Hope you understand.

You know, if EvLer see's this he's gonna' make me do some problems. :yuck:

Last edited: Jun 27, 2005
13. Jun 27, 2005

### Pyrrhus

Shmoe, could you explain the ...? and maybe show a case when it's not possible.

14. Jun 27, 2005

### saltydog

I believe its:

If the sequence of functions:

$$\sum_{n=1}^{\infty}f_n(x)$$

converges uniformly to a function f(x) then:

$$\int_a^b \sum_{n=1}^{\infty} f_n(x)dx=\sum_{n=1}^{\infty} \int_a^b f_n(x)dx$$

If so, then one would need to show uniform convergence for the series:

$$\sum_{n=1}^{\infty}\frac{x^3}{e^{nx}}$$

Someone I hope will correct me if I'm wrong.

15. Jun 27, 2005

### shmoe

$$\int_{0}^{\infty} \sum_{n=1}^{\infty} x^{3}e^{-nx} \ dx=\sum_{n=1}^{\infty} \int_{0}^{\infty} x^{3}e^{-nx} \ dx$$

A series is really just a sequence of functions, so for an example that fails I'll give a sequence of functions where swapping the limit and integral fails. Let $$f_n(x)$$ be 1/n on [0,n] and zero everywhere else. Then $$f_n\rightarrow f(x)$$ where f(x)=0 everywhere (this convergence is even uniform). Now:

$$\int_{0}^{\infty} \lim_{n\rightarrow\infty}f_n(x)dx=\int_{0}^{\infty} f(x)dx=0$$

but

$$\lim_{n\rightarrow\infty}\int_{0}^{\infty} f_n(x) dx=\lim_{n\rightarrow\infty}\int_{0}^{n} \frac{1}{n} dx=\lim_{n\rightarrow\infty} 1 =1$$

Last edited: Jun 27, 2005
16. Jun 28, 2005

### vabamyyr

oh, im sorry saltydog

17. Jun 28, 2005

### saltydog

You know Vabamyyr, someone oughta' tell me: "you know Salty, for not seeing that, for punish work, you need to work through Euler's original computation of $\zeta(2)$. That's right, all the steps. And don't use Mathematica neither!" :surprised

Edit: "and while you're at it, figure out this one too:"

$$\sum_{k=1}^{\infty} \frac{1}{k^{2n}}$$

for all natural numbers n.

Last edited: Jun 28, 2005