Improper Integral

1. Jan 20, 2006

Nusc

Int[(x^3)/((e^x)-1)] [0, infinity]

What is the trick?

I thought of by-parts but how would you integrate 1/((e^x)-1)?

Substitution won't work with
u = e^x -1
with x = ln|u+1|

or it would be rather tough to evaluate u^3

Someone please give me a hint.

2. Jan 20, 2006

StatusX

You can take:

$$\frac{1}{e^x-1} = \frac{e^{-x}}{1-e^{-x}}$$

and then use the substitution u=1-e-x.

3. Jan 20, 2006

Nusc

u=1-e^-x
and dU= e^-x dx
dx = e^x du

e^-x = 1-u
e^x = 1/(1-u)

[(1-u)/u][1/(1-u)] = 1/u

Now putting limits from 0 to infinity.
ln|1-e^-x| = ln(1) - ln(1-1) = 0

That doesn't work.

Last edited: Jan 20, 2006
4. Jan 20, 2006

StatusX

Well, that's how you would get the indefinite integral of 1/(ex-1), which is what I thought you wanted, and although this diverges at 0 as 1/x, your original integral will not because of the x3 in the numerator.

So getting back to the original integral, I don't know how you would find the indefinite integral of this, but the trick to find the integral over the range you specified is this:

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

$$= \int_0^{\infty} x^3 e^{-x} (1+e^{-x}+e^{-2x}+...)dx$$

This can be turned into a sum over the inverse fourth powers of the natural numbers, whose value is, I think, pi^4/90.

5. Jan 21, 2006

benorin

Exactly correct!

Exactly correct! In general: for all real y>1, (or complex y with real part greater than 1,) let

$$I_y=\int_0^{\infty} \frac{x^{y-1} dx}{e^x-1} = \int_0^{\infty} x^{y-1} e^{-x}\frac{1}{1-e^{-x}}dx,$$

expanding the fraction as a geometric series gives

$$\frac{1}{1-e^{-x}}=\sum_{k=0}^{\infty} e^{-kx}$$

and hence

$$I_y= \int_0^{\infty} x^{y-1}e^{-x}\sum_{k=0}^{\infty} e^{-kx}dx = \int_0^{\infty} \sum_{k=1}^{\infty} e^{-kx}x^{y-1}dx = \sum_{k=1}^{\infty} \int_0^{\infty} e^{-kx}x^{y-1} dx$$

substitute $u=kx$ so that $x=\frac{u}{k},$ and hence $dx=\frac{du}{k}$ to get

$$I_y=\sum_{k=1}^{\infty} \int_0^{\infty} e^{-u}\left( \frac{u}{k} \right) ^{y-1} \frac{du}{k} = \sum_{k=1}^{\infty} \left( \frac{1}{k} \right) ^{y} \int_0^{\infty}e^{-u}u^{y-1}du=\zeta (y)\Gamma (y)$$

By the way, I coppied this proof from mathworld.

In particular, we have $$\int_0^{\infty} \frac{x^3 dx}{e^x-1} =\int_0^{\infty} \frac{x^{4-1} dx}{e^x-1} = \zeta (4)\Gamma (4) = \frac{\pi ^4}{90}\cdot 3! = \frac{\pi ^4}{15}$$

Last edited: Jan 21, 2006
6. Jan 22, 2006

Nusc

Is there another and easier way?

I was never introduced to special functions.

7. Jan 22, 2006

StatusX

Not really. All you have to do is get it into a form involving the sum:

$$1+\frac{1}{2^4}+\frac{1}{3^4}+...$$

You can get to this step as long as you know how to do integrals like:

$$\int_0^{\infty} x^3 e^{-a x} dx$$

Once you have it into that form, you can just take it as a given that the above sum is $\pi^4/90$. The fact that pi is involved gives you a clue there probably isn't an easier way, unless there were some lucky trig substitution (which I'm pretty sure there isn't).

8. Nov 15, 2010

khan.muhammad

It was very useful to calculate the Stefan-Boltzmann constant.

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