SOME REPRESENTATIONS OF ζ(3)

Badih Ghusayni

1. Introduction. The Riemann zeta function ζ is defined as

ζ(z) = ∑_{n=1}^{∞} \frac{1}{n^{z}} ,

for each complex number z with real part Re z > 1. In this paper we only concentrate on ζ(3). R. Apéry [1] proved that ζ(3) is an irrational number using the formula

ζ(3) = \frac{5}{2} ∑_{n=1}^{∞} \frac{(−1)^{n−1}}{n^{3}\binom{2n}{n}} .  (1)

Motivated by Apéry’s proof, F. Beukers [2] later gave a shorter proof of the irrationality of ζ(3) by means of double and triple integrals. Beukers’ proof hinged on his formula

ζ(3) = ∫_{0}^{1} ∫_{0}^{1} \frac{− log xy}{1 − xy} dx dy  (2)

where the integrals can be justified by replacing ∫_{0}^{1} with lim_{ε→0} ∫_{ε}^{1−ε}.

The value of ζ(3), however, remains unknown, let alone the values of ζ at other larger odd integers.

The formulas

∑_{n=1}^{∞} \frac{1}{\binom{2n}{n}} = \frac{2π\sqrt{3} + 9}{27},

∑_{n=1}^{∞} \frac{1}{n\binom{2n}{n}} = \frac{π\sqrt{3}}{9},

and

∑_{n=1}^{∞} \frac{1}{n^{2}\binom{2n}{n}} = \frac{π^{2}}{18}

are easy to prove [3]. However, no one knows the value of

∑_{n=1}^{∞} \frac{1}{n^{3}\binom{2n}{n}}.

There is an interesting identity due to Comtet [4] that

∑_{n=1}^{∞} \frac{1}{n^{4}\binom{2n}{n}} = \frac{17π^{4}}{3240}

but there are no known values for

∑_{n=1}^{∞} \frac{1}{n^{k}\binom{2n}{n}}

for integers k > 4.

In section 2 we use Beukers’ formula (2) to find a simple representation of ζ(3) in terms of a single integral instead of a double integral. In section 3 we obtain a series representation for ζ(3). The author hopes that some representation of ζ(3) in the literature can be used to evaluate ζ(3).

2. An Integral Representation of ζ(3). Let us write Beukers’ formula as

ζ(3) = \frac{1}{2} ∫_{0}^{1} ∫_{0}^{1} \frac{log xy}{xy − 1} dx dy

and consider (x, y) ∈ (0, 1) × (0, 1).

For a fixed y, substitute w = xy − 1 in the innermost integral. Then

ζ(3) = \frac{1}{2} ∫_{0}^{1} \frac{1}{y} ∫_{−1}^{y−1} \frac{log(w + 1)}{w} dw dy = \frac{1}{2} ∫_{0}^{1} \frac{1}{y} lim_{a→(−1)^{+}} ∫_{a}^{y−1} \frac{log(w + 1)}{w} dw dy.

Now

log(w + 1) = ∑_{n=1}^{∞}(−1)^{n+1} \frac{w^{n}}{n} , |w| ≤ 1.

Then

∑_{n=1}^{∞}(−1)^{n+1} \frac{w^{n−1}}{n}

converges uniformly to \frac{log(w+1)}{w} on [a, y − 1]. So

ζ(3) = \frac{1}{2} ∫_{0}^{1} \frac{1}{y} lim_{a→(−1)^{+}} ∑_{n=1}^{∞} \frac{(−1)^{n+1}}{n} ∫_{a}^{y−1} w^{n−1} dwdy

= \frac{1}{2} ∫_{0}^{1} \frac{1}{y} ∑_{n=1}^{∞} \frac{(−1)^{n+1}}{n} \{\frac{(y − 1)^{n}}{n} − \frac{(−1)^{n}}{n} \} dy

= \frac{1}{2} ∫_{0}^{1} \frac{1}{y} ∑_{n=1}^{∞}\{\frac{(−1)^{n+1}(y − 1)^{n}}{n^{2}} + \frac{1}{n^{2}} \} dy

= \frac{1}{2} ∫_{0}^{1} \frac{1}{y} \{∑_{n=1}^{∞} \frac{(−1)^{n+1}(y − 1)^{n}}{n^{2}} + \frac{π^{2}}{6} \} dy

since y − 1 ∈ (0, 1) and absolute convergence implies convergence. Using the functional equation for the dilogarithm

∑_{n=1}^{∞} \frac{y^{n}}{n^{2}}

[3] we have

∑_{n=1}^{∞} \frac{(−1)^{n+1}(y − 1)^{n}}{n^{2}} = − ∑_{n=1}^{∞} \frac{(1 − y)^{n}}{n^{2}}

= log(1 − y) log y + ∑_{n=1}^{∞} \frac{y^{n}}{n^{2}} − \frac{π^{2}}{6}

so

ζ(3) = \frac{1}{2} ∫_{0}^{1} \frac{1}{y} \{log(1 − y) log y + ∑_{n=1}^{∞} \frac{y^{n}}{n^{2}} \} dy

= \frac{1}{2} ∫_{0}^{1} \frac{log(1 − y) log y}{y} dy + \frac{1}{2} ∑_{n=1}^{∞} \frac{1}{n^{3}}

or

ζ(3) = ∫_{0}^{1} \frac{log(1 − y) log y}{y} dy.

It is worth mentioning that, by making a simple change of variable, the above integral representation can be written as

ζ(3) = ∫_{0}^{1} \frac{log x}{x − 1} log \frac{1}{1 − x} dx

where it is easy to see that

∫_{0}^{1} \frac{log x}{x − 1} dx = \frac{π^{2}}{6} = ζ(2).

3. A Series Representation of ζ(3). Using the well–known formula [3]

2(sin^{−1} x)^{2} = ∑_{n=1}^{∞} \frac{(2x)^{2n}}{n^{2}\binom{2n}{n}}

we have

2 ∫_{0}^{1/2}(sin^{−1} y)^{2} \frac{dy}{y} = \frac{1}{2} ∑_{n=1}^{∞} \frac{1}{n^{3}\binom{2n}{n}}.

On the other hand a simple integration by substitution followed by integration by parts yields

2 ∫_{0}^{1/2}(sin^{−1} y)^{2} \frac{dy}{y} = − ∫_{0}^{π/3} x log(2 sin \frac{1}{2}x)dx.

Combining we get

∑_{n=1}^{∞} \frac{1}{n^{3}\binom{2n}{n}} = −2 ∫_{0}^{π/3} x log(2 sin \frac{1}{2}x)dx.  (3)

Now clearly

log(1 − z) = − ∑_{n=1}^{∞} \frac{z^{n}}{n} , |z| < 1

even at boundary points except for z = 1, i.e. except at the points z = e^{ix} with x ≠ 2kπ. Consider the interval (0, 2π). Now

1 − z = 1 − e^{ix} = (1 − cos x) − sin xi

= 2 sin^{2} \frac{x}{2} − 2 sin \frac{x}{2} cos \frac{x}{2}i

= 2 sin \frac{x}{2}(sin \frac{x}{2} − cos \frac{x}{2}i)

= 2 sin \frac{x}{2}\{cos(−\frac{π}{2} + \frac{x}{2}) + sin(−\frac{π}{2} + \frac{x}{2})i\}

= 2 sin \frac{x}{2}e^{(− \frac{π}{2} + \frac{x}{2} )i}.

So

log(1 − z) = log(2 sin \frac{x}{2}) + (−\frac{π}{2} + \frac{x}{2})i.

Applying Abel’s theorem for trigonometric series we get

log(2 sin \frac{x}{2}) = − ∑_{n=1}^{∞} \frac{cos nx}{n}, x ∈ (0, 2π).

Using formula (3) we can now write

∑_{n=1}^{∞} \frac{1}{n^{3}\binom{2n}{n}} = 2 ∑_{n=1}^{∞} \frac{1}{n} ∫_{0}^{π/3} x cos nx dx.

Integrating by parts twice we get

∑_{n=1}^{∞} \frac{1}{n^{3}\binom{2n}{n}} = \frac{2π}{3} ∑_{n=1}^{∞} \frac{sin \frac{nπ}{3}}{n^{2}} + 2 ∑_{n=1}^{∞} \frac{cos \frac{nπ}{3}}{n^{3}} − 2ζ(3).

So

ζ(3) = \frac{π}{3} ∑_{n=1}^{∞} \frac{sin \frac{nπ}{3}}{n^{2}} + ∑_{n=1}^{∞} \frac{cos \frac{nπ}{3}}{n^{3}} − \frac{1}{2} ∑_{n=1}^{∞} \frac{1}{n^{3}\binom{2n}{n}}.

Since the middle term is \frac{1}{3}ζ(3) [5], we consequently have the following series representation

ζ(3) = \frac{π}{2} ∑_{n=1}^{∞} \frac{sin \frac{nπ}{3}}{n^{2}} − \frac{3}{4} ∑_{n=1}^{∞} \frac{1}{n^{3}\binom{2n}{n}}.

\underline{References}

1. R. Apéry, “Irrationalité de ζ(2) et ζ(3),” Journées Arithmétiques de Luminy, Astérisque, 61 (1979), 11–13.

2. F. Beukers, “A Note on the Irrationality of ζ(2) and ζ(3),” Bull. London Math. Soc., 11 (1979), 268–272.

3. J. Borwein and P. Borwein, Pi and the AGM, Wiley–Interscience, New York, 1987.

4. L. Comtet, Advanced Combinatorics, Dreidel, Dordrecht, 1974.

5. L. Lewin, Polylogarithms and Associated Functions, North–Holland, New York, 1981.

Badih Ghusayni Department of Mathematics and Statistics Notre Dame University P.O. Box: 72 Zouk Mikhael Zouk Mosbeh, Lebanon email: badih.j@dm.net.lb