# Dirichlet eta function: Wikis

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# Encyclopedia

Dirichlet eta function η(s) in the complex plane. The color of a point s encodes the value of η(s). Strong colors denote values close to zero and hue encodes the value's argument.

In mathematics, in the area of analytic number theory, the Dirichlet eta function can be defined as

$\eta(s) = \left(1-2^{1-s}\right) \zeta(s)$

where ζ is Riemann's zeta function. However, it can also be used to define the zeta function. It has a Dirichlet series expression, valid for any complex number s with positive real part, given by

$\eta(s) = \sum_{n=1}^{\infty}{(-1)^{n-1} \over n^s}.$

While this is convergent only for s with positive real part, it is Abel summable for any complex number, which serves to define the eta function as an entire function, and shows the zeta function is meromorphic with a single pole at s = 1.

Equivalently, we may begin by defining

$\eta(s) = \frac{1}{\Gamma(s)}\int_0^\infty \frac{x^{s-1}}{e^x+1}{dx}$

which is also defined in the region of positive real part. This gives the eta function as a Mellin transform.

Hardy gave a simple proof of the functional equation for the eta function, which is

$\eta(-s) = 2\pi^{-s-1} s \sin\left({\pi s \over 2}\right) \Gamma(s)\eta(s+1).$

From this, one immediately has the functional equation of the zeta function also, as well as another means to extend the definition of eta to the entire complex plane.

## Numerical Algorithms

Most of the series acceleration techniques developed for alternating series can be profitably applied to the evaluation of the eta function. One particularly simple, yet reasonable method is to apply Euler's transformation of alternating series, to obtain

$\eta(s)=\sum_{n=0}^\infty \frac{1}{2^{n+1}} \sum_{k=0}^n (-1)^{k} {n \choose k} \frac {1}{(k+1)^s}.$

Note that the second, inside summation is a forward difference.

### Borwein's method

Peter Borwein used approximations involving Chebyshev polynomials to produce a method for efficient evaluation of the eta function. If

$d_k = n\sum_{i=0}^k \frac{(n+i-1)!4^i}{(n-i)!(2i)!}$

then

$\eta(s) = -\frac{1}{d_n} \sum_{k=0}^{n-1}\frac{(-1)^k(d_k-d_n)}{(k+1)^s}+\gamma_n(s),$

where for $\Re(s) \ge \frac{1}{2}$ the error term γn is bounded by

$|\gamma_n(s)| \le \frac{3}{(3+\sqrt{8})^n} (1+2|\Im(s)|)\exp(\frac{\pi}{2}|\Im(s)|).$

The factor of $3+\sqrt{8}\approx 5.8$ in the error bound indicates that the Borwein series converges quite rapidly as n increases.

## Particular values

Also:

$\!\ \eta(1) = \ln2$, this is the alternating harmonic series
$\eta(2) = {\pi^2 \over 12}$
$\eta(4) = {{7\pi^4} \over 720}$
$\eta(6) = {{31\pi^6} \over 30240}$
$\eta(8) = {{127\pi^8} \over 1209600}$
$\eta(10) = {{73\pi^{10}} \over 6842880}$
$\eta(12) = {{1414477\pi^{12}} \over {1307674368000}}$

The general form for even positive integers is:

$\eta(2n) = (-1)^{n+1}{{B_{2n}\pi^{2n}(2^{2n-1} - 1)} \over {(2n)!}}.$