$\int_{0}^{\infty} [x^{3} / (e^{x} - 1)] dx$

By Arun Iyer on Wednesday, December 19, 2001 - 06:13 pm:

could anyone please evaluate....

\int_{0}^{\infty} [x^{3} / (e^{x} - 1)] dx

love arun

By Michael Doré on Thursday, December 20, 2001 - 01:04 am:

I would suggest splitting up $x^{3} / (e^{x} - 1)$ as:

$x^{3} e^{- x} / (1 - e^{- x}) = x^{3} (e^{- x} + e^{- 2 x} + e^{- 3 x} + . . .)$

i.e. a geometric series (which we can do since $e^{- x} < 1$ for $x > 0$ ).

Note that $\int_{0}^{\infty} x^{3} e^{- n x} dx = \int_{0}^{\infty} u^{3} e^{- u} / n^{4} du = 6 / n^{4}$ .

So the answer is $6 / 1^{4} + 6 / 2^{4} + . . . = 6 ζ (4)$ . I can't actually remember the value of $ζ (4)$ offhand, but it is well known (it is a rational multiple of $π^{4}$ ).

By Arun Iyer on Thursday, December 20, 2001 - 05:58 pm:

Well, that's some neat transformation there...beautiful.

I know that

ζ (4) = π^{4} / 90

However, do you know a way of proving this...Michael?

love arun

By Arun Iyer on Saturday, December 22, 2001 - 06:42 pm:
Well, anyone know proof for

ζ (4) = π^{4} / 90

?
love arun

By Yatir Halevi on Saturday, December 22, 2001 - 07:14 pm:

I know that it is not a proof
but i did see somewhere that:

ζ (2 k) = 2^{2 k - 1} π^{2 k} B_{k} / (2 k)!

$k = 1$ , $2$ , $3$ , ...

let's take $k = 2$ .

we get:

The second bernoulli number is 1/30

$2^{3} π^{4} / 30 \times 24 = π^{4} / 90$
Yatir

By Arun Iyer on Saturday, December 22, 2001 - 07:36 pm:

i know that formula...
i had given the link for that some months back..

this is the site..
http://numbers.computation.free.fr/Constants/Miscellaneous/bernoulli.html

love arun

By Michael Doré on Sunday, December 23, 2001 - 01:44 am:

I think it is easy if you're willing to accept Euler's sine product. If we start with:

$\sin x = x (1 - x^{2} / π^{2}) (1 - x^{2} / (4 π^{2})) . . .$

then on taking logarithms and differentiating you get:

$\cot x = 1 / x + \sum [1 / (x + r π) + 1 / (x - r π)]$

where $r$ is taken over the naturals.

Differentiate each side $2 n - 1$ times:

$d^{2 n - 1} / {dx}^{2 n - 1} \cot x = - (2 n - 1)! (1 / x^{2 n} + \sum 1 / (x + r π)^{2 n} + \sum 1 / (x - r π)^{2 n})$

Bring the $1 / x^{2 n}$ term to the other side then take the limit as $x \to 0$ :

$lim_{x \to 0} ((d^{2 n - 1} / {dx}^{2 n - 1} \cot x) + (2 n - 1)! / x^{2 n}) = - 2 (2 n - 1)! ζ (2 n) / π^{2 n}$

Write $\cot x = i (e^{2 i x} + 1) / (e^{2 i x} - 1)$ , multiply throughout by $- i$ and you get:

$lim_{x \to 0} (d^{2 n - 1} / {dx}^{2 n - 1} ((e^{2 i x} + 1) / (e^{2 i x} - 1)) - i (2 n - 1)! / x^{2 n}) = 2 i (2 n - 1)! ζ (2 n)$

So:

$lim_{x \to 0} (d^{2 n - 1} / {dx}^{2 n - 1} ((e^{2 i x} + 1) / (2 i x) \times 2 i x / (e^{2 i x} - 1)) - i (2 n - 1)! / x^{2 n}) = 2 i (2 n - 1)! ζ (2 n)$

Setting $y = 2 i x$ and using the chain rule this becomes:

$lim_{y \to 0} (d^{2 n - 1} / {dy}^{2 n - 1} ((e^{y} + 1) / y \times y / (e^{y} - 1)) \times 2^{2 n - 1} (- 1)^{n + 1} i + 2^{2 n} (- 1)^{n + 1} (2 n - 1)! i / y^{2 n}) = 2 i (2 n - 1)! ζ (2 n)$ (*)

Note that $(e^{y} + 1) / y$ and $y / (e^{y} - 1)$ can both be series expanded:

$(e^{y} + 1) / y = 2 / y + 1 + y / 2! + y^{2} / 3! + y^{3} / 4!$

$y / (e^{y} - 1) = B_{0} + B_{1} y + B_{2} y^{2} / 2! + . . .$

where $B_{n}$ is the $n$ th Bernoulli number (using the well known fact that $y / (e^{y} - 1)$ is the generating function for the Bernoulli numbers).

If you substitute these series in (*) and expand out the product then you should find that the reciprocal terms cancel (so luckily the function doesn't blow up at the origin) and then the final result follows on applying the (well known??) identity:

$n! B_{n} = \sum_{r = 1}^{n} C (n, r) B_{r}$

which is easily proved by equating coefficients in the identity:

$x / (e^{x} - 1) \times e^{x} = x + x / (e^{x} - 1)$

$(e^{x} + 1) / (e^{2 x} - 1) = 1 / (e^{x} - 1)$ .

Of course this can all be greatly speeded up for $ζ (4)$ .

By Michael Doré on Sunday, December 23, 2001 - 02:56 am:

Oops, ignore the penultimate line. And I think the identity is only for n > 1.

By Arun Iyer on Sunday, December 23, 2001 - 05:47 pm:

whoa!!pretty long proof....phew!!

thanks michael...