Editing Dirichlet eta function (section)

==Landau's problem with ''ζ''(''s'') = ''η''(''s'')/ (1 − (2^(1−s)))  and solutions==
In the equation {{math|1=''η''(''s'') = (1 − 2<sup>1−''s''</sup>) ''ζ''(''s'')}}, "the pole of {{math|''ζ''(''s'')}} at {{math|1=''s'' = 1}} is cancelled by the zero of the other factor" (Titchmarsh, 1986, p.&nbsp;17), and as a result {{math|''η''(1)}} is neither infinite nor zero (see {{section link||Particular values}}). However, in the equation
<math display="block">\zeta(s)=\frac{\eta(s)}{1-2^{1-s}},</math>
''η'' must be zero at all the points <math>s_n = 1+n\frac{2\pi}{\ln{2}}i, n\ne0, n \in \Z </math>, where the denominator is zero, if the Riemann zeta function is analytic and finite there. The problem of proving this without defining the zeta function first was signaled and left open by [[Edmund Landau|E. Landau]] in his 1909 treatise on number theory: "Whether the eta series is different from zero or not at the points <math>s_n\ne1</math>, i.e., whether these are poles of zeta or not, is not readily apparent here."

A first solution for Landau's problem was published almost 40 years later by [[David Widder|D. V. Widder]] in his book The Laplace Transform. It uses the next prime 3 instead of 2 to define a Dirichlet series similar to the eta function, which we will call the <math>\lambda</math> function, defined for <math>\Re(s)>0</math> and with some zeros also on <math>\Re(s) = 1</math>, but not equal to those of eta.
{{math proof|title=Indirect proof of {{math|1=''η''(''s''<sub>''n''</sub>) = 0}} following Widder|proof=
<math display="block">\lambda(s)
= \left(1-\frac{3}{3^s}\right)\zeta(s)
= \left(1+\frac{1}{2^s}\right)-\frac{2}{3^s}+\left(\frac{1}{4^s}+\frac{1}{5^s}\right)-\frac{2}{6^s} + \cdots</math>

If <math>s</math> is real and strictly positive, the series converges since the regrouped terms alternate in sign and decrease in absolute value to zero. According to a theorem on uniform convergence of Dirichlet series first proven by Cahen in 1894, the <math>\lambda(s)</math> function is then analytic for <math>\Re(s)>0</math>, a region which includes the line 
<math>\Re(s)=1</math>. Now we can define correctly, where the denominators are not zero,
<math display="block">\zeta(s) = \frac{\eta(s)}{1-\frac{2}{2^s}}</math>
or
<math display="block">\zeta(s) = \frac{\lambda(s)}{1-\frac{3}{3^s}}</math>

Since <math>\frac{\log 3}{\log 2}</math> is irrational, the denominators in the two definitions are not zero at the same time except for <math>s=1</math>, and the <math>\zeta(s)\,</math> function is thus well defined and analytic for <math>\Re(s)>0</math> except at <math>s=1</math>. We finally get indirectly that <math>\eta(s_n)=0</math> when <math>s_n\ne1</math>:
<math display="block">
\eta(s_n) = \left(1-\frac{2}{2^{s_n}}\right)\zeta(s_n)
          = \frac{1-\frac{2}{2^{s_n}}}{1-\frac{3}{3^{s_n}}} \lambda(s_n) = 0.
</math>
}}

An elementary direct and <math>\zeta\,</math>-independent proof of the vanishing of the eta function at <math>s_n\ne1</math> was published by J. Sondow in 2003. It expresses the value of the eta function as the limit of special  Riemann sums associated to an integral known to be zero, using a relation between the partial sums of the Dirichlet series defining the eta and zeta functions for <math>\Re(s)>1</math>.

{{math proof|title=Direct proof of {{math|1=''η''(''s''<sub>''n''</sub>) = 0}} by Sondow|proof=
With some simple algebra performed on finite sums, we can write for any complex ''s''
<math display="block">\begin{align}
\eta_{2n}(s)
&=   \sum_{k=1}^{2n}\frac{(-1)^{k-1}}{k^s} \\
&=   1-\frac{1}{2^s}+\frac{1}{3^s}-\frac{1}{4^s}+\dots+\frac{(-1)^{2n-1}}{{(2n)}^s} \\[2pt]
&=   1+\frac{1}{2^s}+\frac{1}{3^s}+\frac{1}{4^s}+\dots+\frac{1}{{(2n)}^s}
  - 2\left(\frac{1}{2^s}+\frac{1}{4^s}+\dots+\frac{1}{{(2n)}^s}\right) \\[2pt]
&=   \left(1-\frac{2}{2^s}\right)\zeta_{2n}(s) + \frac{2}{2^s}\left(\frac{1}{{(n+1)}^s} + \dots + \frac{1}{{(2n)}^s}\right) \\[2pt]
&=   \left(1-\frac{2}{2^s}\right)\zeta_{2n}(s) + \frac{2n}{{(2n)}^s}\,\frac{1}{n}\,\left(\frac{1}{{(1+1/n)}^s}+ \dots + \frac{1}{{(1+n/n)}^s} \right).
\end{align}</math>

Now if <math>s=1+it</math> and <math>2^s=2</math>, the factor multiplying <math>\zeta_{2n}(s)</math> is zero, and
<math display="block">\eta_{2n}(s) = \frac{1}{n^{it}} R_n\left(\frac{1}{{(1+x)}^s},0,1\right),</math>
where {{math|''R''<sub>''n''</sub>(''f''(''x''), ''a'', ''b'')}} denotes a special Riemann sum approximating the integral of {{math|''f''(''x'')}} over {{closed-closed|''a'', ''b''}}.
For {{math|1=''t'' = 0}} i.e., {{math|1=''s'' = 1}}, we get
<math display="block">\eta(1) = \lim_{n\to\infty} \eta_{2n}(1) = \lim_{n\to\infty} R_n\left(\frac{1}{1+x},0,1\right) = \int_0^1 \frac{dx}{1+x} = \log 2 \ne 0.</math>

Otherwise, if <math>t \ne 0</math>, then <math>|n^{1-s}|=|n^{-it}|=1</math>, which yields
<math display="block">
|\eta(s)| = \lim_{n\to\infty} |\eta_{2n}(s)| = \lim_{n\to\infty} \left|R_n\left(\frac{1}{{(1+x)}^s},0,1\right)\right|
   =  \left|\int_0^1 \frac{dx}{{(1+x)}^s}\right| = \left|\frac{2^{1-s}-1}{1-s}\right| = \left|\frac{1-1}{-it}\right| = 0.
</math>
}}

Assuming <math>\eta(s_n)=0</math>, for each point <math>s_n\ne1</math> where <math>2^{s_n}=2</math>, we can now define <math>\zeta(s_n)\,</math> by continuity as follows,
<math display="block">
\zeta(s_n) = \lim_{s\to s_n}\frac{\eta(s)}{1-\frac{2}{2^s}}
 = \lim_{s\to s_n}\frac{\eta(s)-\eta(s_n)}{\frac{2}{2^{s_n}}-\frac{2}{2^s}}
 = \lim_{s\to s_n}\frac{\eta(s)-\eta(s_n)}{s-s_n}\,\frac{s-s_n}{\frac{2}{2^{s_n}}-\frac{2}{2^s}}
 = \frac{\eta'(s_n)}{\log(2)}.
</math>

The apparent singularity of zeta at <math>s_n\ne1</math> is now removed, and the zeta function is proven to  be analytic everywhere in <math>\Re{s} > 0</math>, except at <math>s=1</math> where
<math display="block">
\lim_{s \to 1} (s-1)\zeta(s) = \lim_{s\to 1} \frac{\eta(s)}{\frac{1-2^{1-s}}{s-1}}
 = \frac{\eta(1)}{\log 2} = 1.
</math>

==Integral representations==
A number of integral formulas involving the eta function can be listed. The first one follows from a change of variable of the integral representation of the Gamma function (Abel, 1823), giving a [[Mellin transform]] which can be expressed in different ways as a double integral (Sondow, 2005). This is valid for <math>\Re s > 0.</math>
<math display="block">\begin{align}
\Gamma(s)\eta(s)
&= \int_0^\infty \frac{x^{s-1}}{e^x+1} \, dx
 = \int_0^\infty \int_0^x \frac{x^{s-2}}{e^x+1} \, dy \, dx \\[8pt]
&= \int_0^\infty \int_0^\infty \frac{(t+r)^{s-2}}{e^{t+r}+1} dr \, dt
 =\int_0^1 \int_0^1 \frac{\left(-\log(x y)\right)^{s-2}}{1 + x y} \, dx \, dy.
\end{align}</math>

The Cauchy–Schlömilch transformation (Amdeberhan, Moll et al., 2010) can be used to prove this other representation, valid for <math>\Re s > -1</math>. [[Integration by parts]] of the first integral above in this section yields another derivation.

<math display="block">2^{1-s}\,\Gamma(s+1)\,\eta(s)
= 2 \int_0^\infty \frac{x^{2s+1}}{\cosh^2(x^2)} \, dx
= \int_0^\infty \frac{t^s}{\cosh^2(t)} \, dt.</math>

The next formula, due to Lindelöf (1905), is valid over the whole complex plane, when the principal value is taken for the logarithm implicit in the exponential.
<math display="block">\eta(s) = \int_{-\infty}^\infty \frac{(1/2 + i t)^{-s}}{e^{\pi t}+e^{-\pi t}} \, dt.</math>
This corresponds to a [[Johan Jensen (mathematician)|Jensen]] (1895) formula for the entire function 
<math>(s-1)\,\zeta(s)</math>, valid over the whole complex plane and also proven by Lindelöf.
<math display="block">(s-1)\zeta(s) = 2\pi\,\int_{-\infty}^\infty \frac{(1/2 + i t)^{1-s}}{(e^{\pi t}+e^{-\pi t})^2} \, dt. </math>
"This formula, remarquable by its simplicity, can be proven easily with the help of Cauchy's theorem, so important for the summation of series" wrote Jensen (1895). Similarly by converting the integration paths to contour integrals one can obtain other formulas for the eta function, such as this generalisation (Milgram, 2013) valid for <math>0 < c < 1</math> and all <math>s</math>:
<math display="block">\eta(s) = \frac{1}{2} \int_{-\infty}^\infty \frac{(c + i t)^{-s}}{\sin{(\pi(c+i t))}} \, dt.</math>
The zeros on the negative real axis are factored out cleanly by making <math>c\to 0^+</math> (Milgram, 2013) to obtain a formula valid for <math>\Re s < 0</math>:
<math display="block">\eta(s) = - \sin\left(\frac{s\pi}{2}\right) \int_{0}^\infty \frac{t^{-s}}{\sinh{(\pi t)}} \, dt.</math>