Evaluating Integral: \int_0^\infty\frac{x^{3}}{e^x-1}dx

[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?

 

[dextercioby]

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

Daniel.

 

[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

 

[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 as well 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 $$\frac{1}{e^{x}-1}$$

 

[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.

 

[vabamyyr]

it is getting over my head, i think ill have to learn more before doing that task

 

[dextercioby]

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

is all you need.

Daniel.

 

[vabamyyr]

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

 

[saltydog]

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.

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:

$$\Gamma(4)=6$$

 

[vabamyyr]

there are several shapes to express zeta function and u don't have to tell me what gamma function is

 

[shmoe]

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

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.

 

[saltydog]

there are several shapes to express zeta function and u don't have to tell me what gamma function is

Vabamyyr, sorry you mis-understood my comments: I was saying that to me. 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.

 

[Pyrrhus]

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.

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

 

[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.

 

[shmoe]

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

is the step I'm talking about. There was a recent post about dominated convergence if you want some details about this.

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

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

 

[vabamyyr]

oh, I am sorry saltydog

 

[saltydog]

oh, I am sorry 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!"

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.