What is the symmetric functional equation of the Dirichlet eta function? At the page fractional calculus, completed Riemann zeta
it is claimed that the symmetric functional equation for the Dirichlet eta function is:
Formula 7.3.2
$$\Gamma\left(\frac{z}{2}\right)\pi^{\Large-\frac{z}{2}}(1-2^z)\eta(z)=\Gamma\left(\frac{1-z}{2}\right)\pi^{\Large-\frac{1-z}{2}}(1-2^{1-z})\eta(1-z) \;\;\;\;\;\;\;(1)$$
Where $z \neq 0,1 $
and the following equation that I have not checked:
$$\pi^{\Large-\frac{z}{2}}\Gamma\left\{\frac{1}{2}\left(\frac{1}{2}+z\right)\right\}\left(1-2^{\Large\frac{1}{2}+z}\right)\eta\left(\frac{1}{2}+z\right)=\pi^{\Large\frac{z}{2}}\Gamma\left\{\frac{1}{2}\left(\frac{1}{2}-z\right)\right\}\left(1-2^{\Large\frac{1}{2}-z}\right)\eta\left(\frac{1}{2}-z\right) \;\;\;\;\;\;\;\;\;(2)$$
Where $z \neq \pm \frac{1}{2} $
But is the first equation really correct? Is the Dirichlet eta function completed in this way? I am new to the symmetric functional equation for the Riemann zeta function but to my mind this seems to be equivalent to completing the zeta function as:
$$\zeta(1-z)\zeta(z)$$
which other mathematicians would say is not the way to complete the zeta function.
What is the correct way to complete the Dirichlet eta function?
$$\eta(s)=\zeta(s)\left(1-1/2^{s-1}\right)$$
 A: I don't understand your objection; there's nothing like $\zeta(1 - s) \zeta(s)$ appearing here. (Also, it is customary to use $s$ to denote a complex variable, not $z$, when referring to zeta functions.)
The completed zeta function
\[\Lambda(s) = \pi^{-s/2} \Gamma\left(\frac{s}{2}\right) \zeta(s)\]
satisfies the functional equation
\[\Lambda(1 - s) = \Lambda(s)\]
for all $s \in \mathbb{C} \setminus \{0,1\}$. As
\[\eta(s) = (1 - 2^{1 - s}) \zeta(s),\]
it follows that
\[\Lambda(1 - s) = \pi^{-(1-s)/2} \Gamma\left(\frac{1 - s}{2}\right) \zeta(1 - s) = \pi^{-(1-s)/2} \Gamma\left(\frac{1 - s}{2}\right) (1 - 2^s)^{-1} \eta(1 - s) \]
and that
\[\Lambda(s) = \pi^{-s/2} \Gamma\left(\frac{s}{2}\right) \zeta(s) = \pi^{-s/2} \Gamma\left(\frac{s}{2}\right) (1 - 2^{1 - s})^{-1} \eta(s),\]
and these are both equal by the functional equation.
If you want the symmetric functional equation (though I can't imagine why you would), then you just replace $s$ with $1/2 + s$. The result, after some simple rearranging, is
\[\pi^{-\frac{1}{2} \left(\frac{1}{2} - s\right)} \Gamma\left(\frac{1}{2} \left(\frac{1}{2} - s\right)\right) \left(1 - 2^{\frac{1}{2} - s}\right) \eta\left(\frac{1}{2} - s\right) = \pi^{-\frac{1}{2} \left(\frac{1}{2} + s\right)} \Gamma\left(\frac{1}{2} \left(\frac{1}{2} + s\right)\right) \left(1 - 2^{\frac{1}{2} + s}\right) \eta\left(\frac{1}{2} + s\right).\]
Again, this is all follows trivially from the functional equation for the Riemann zeta function and some trivial rearranging.
