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:


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?



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.

  • $\begingroup$ I am interested in the equations: $$s=s$$ $$\pi ^{-\frac{s}{2}} \left(1-2^s\right) \zeta (s) \left(1-\frac{1}{2^{s-1}}\right) \Gamma \left(\frac{s}{2}\right)$$ with the substitution: $$s=1-s$$ $$\pi ^{-\frac{s}{2}} \left(1-2^s\right) \zeta (s) \left(1-\frac{1}{2^{s-1}}\right) \Gamma \left(\frac{s}{2}\right)$$ $\endgroup$ – Mats Granvik Aug 17 '16 at 16:02
  • $\begingroup$ What about these equations? $\endgroup$ – Peter Humphries Aug 17 '16 at 16:05
  • $\begingroup$ I want to apply this to: math.stackexchange.com/questions/1832268/… and: math.stackexchange.com/questions/1893891/… and then run through the Franca LeClair derivation of the asymptotic for the Riemann zeta zeros: described here: arxiv.org/abs/1407.4358 at page 34. $\endgroup$ – Mats Granvik Aug 17 '16 at 16:10
  • $\begingroup$ Do bear in mind that all of Franca and LeClair's results aren't rigorous (there is a reason that these papers are just preprints and not peer-reviewed publications). $\endgroup$ – Peter Humphries Aug 17 '16 at 16:14
  • $\begingroup$ Yes but the numerical evidence for their asymptotic is good when checking for the first 100000 zeros. Anyways, what they don't state clearly in their paper is that their asymptotic is the nearest integer to what could be called the "complementary" Gram points here in the OEIS: oeis.org/A273061 and not the nearest integer to the zeta zeros. $\endgroup$ – Mats Granvik Aug 17 '16 at 16:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.