# Improper Integral trick

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