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I am trying to determine in what way to approach finding a connection between Dedekind's Eta Function, defined as $$\eta(\tau)=q^\frac{1}{24}\prod_{n=1}^\infty(1-q^n)$$ where $q=e^{2\pi i \tau}$ is referred to as the nome.

and the Gamma Function $$\Gamma(s)=\int_{0}^\infty x^{s-1}e^{-x}dx$$ More specifically I would like to understand through what methods these identities are derived: $$\eta(i)=\Gamma(\frac{1}{4})\frac{\pi^{-3/4}}{2}$$ $$\eta(2i)=\Gamma(\frac{1}{4})2^{-11/8}\pi^{-3/4}$$ And in general what seems to be $$\eta(ki)=\Gamma(\frac{1}{4})\pi^{-3/4}C_{k}$$ for whole numbers $k$ and some constant $C_k$ Where $C_k$ looks to be algebraic for $k\in 1,2,3,4$. I guess what I really want to know is why does this factor of $\Gamma(\frac{1}{4})\pi^{-3/4}$ come into play at imaginary integer values for the $\eta$ function?

I know there is a relationship between the $\eta$ and Jacobi Theta Functions that can be found using the Pentagonal Number Theorem or Jacobi's Triple Product Identity but I do not know how it fits into evaluation of $\eta(ki)$.

EDIT:My attempt at an answer: $$\int_{-\infty}^\infty e^{-x^{2p}} dx=\frac{\Gamma(\frac{1}{2p})}{p}$$ can be derived through substitution. $$\frac{\Gamma^2(\frac{1}{2p})}{p^2}=\int_\Bbb {R^2}\exp(-(x^{2p}+y^{2p})dxdy$$ Applying the coordinate transformation $x^{2p}+y^{2p}=r^{2p}$ with $x=r\frac{\cos(\phi)}{|\sin(\phi)|^{2p}+|cos(\phi)|^{2p}}$ and $y=r\frac{\sin(\phi)}{|\sin(\phi)|^{2p}+|cos(\phi)|^{2p}}$ I get$$\frac{\Gamma^2(\frac{1}{2p})}{p^2}=\int_{0}^\infty re^{-r^{2p}}dr\int_{0}^{2\pi}\frac{d\phi}{(\sin^{2p}(\phi)+\cos^{2p}(\phi))^{\frac{1}{p}}}$$ The integral over $r$ evaluates to $\frac{\Gamma(\frac{1}{p})}{2p}$

So$$\frac{2\Gamma^2(\frac{1}{2p})}{p\Gamma({\frac{1}{p})}}=\int_{0}^{2\pi}\frac{d\phi}{(\sin^{2p}(\phi)+\cos^{2p}(\phi))^{\frac{1}{p}}}$$ The integral is symmetric over $[0,\pi]$ and $[\pi, 2\pi]$ so we get $$\frac{\Gamma^2(\frac{1}{2p})}{p\Gamma({\frac{1}{p})}}=\int_{0}^{\pi}\frac{d\phi}{(\sin^{2p}(\phi)+\cos^{2p}(\phi))^{\frac{1}{p}}}$$ Plugging in $p=2$ yields $$\frac{\Gamma^2(\frac{1}{4})}{2\sqrt{\pi}}=\int_{0}^\pi \frac{d\phi}{\sqrt{\sin^4(\phi)+\cos^4(\phi)}}$$Using $u=\cos(\phi)$ I arrive at $$\frac{\Gamma^2(\frac{1}{4})}{2\sqrt{\pi}}=\int_{-1}^1 \frac{du}{\sqrt{(2u^4-2u^2+1)(1-u^2)}}$$

$$\frac{\Gamma^2(\frac{1}{4})}{4\sqrt{\pi}}=\int_{0}^1 \frac{du}{\sqrt{-2u^6+5u^4-3u^2+1}}$$ This looks to be similar to an elliptic integral but I am finding trouble reducing it to a form that I can evaluate.

EDIT: If I can evaluate the integral in terms of the Complete Elliptic Integral of the First Kind, I can use its relation with Jacobi's Third Theta Function to evaluate it in terms of $\eta$. Such that$$\frac{\Gamma^2(\frac{1}{4})}{4\sqrt{\pi}}=cK(k')=\frac{\pi}{2}\theta_3^2(q)$$ So that we arrive at the familiar form on the LHS $$\frac{\Gamma(\frac{1}{4})\pi^{-3/4}}{2}=\frac{\theta_3(q)}{\sqrt{2c}}$$

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  • $\begingroup$ This is a possible duplicate of : math.stackexchange.com/questions/1334684/… $\endgroup$ – Mason Jul 16 '18 at 22:24
  • $\begingroup$ @Mason I don't see how my question is a duplicate. The other OP asks for the value of $\eta(6i)$ whereas I want to know why the ratio $\Gamma(1/4)\pi^{-3/4}$ appears in those values. The other question did not make mention of this at all. $\endgroup$ – aleden Jul 16 '18 at 23:02
  • $\begingroup$ The question does make mention of this? It is question 2 of his 2 questions. But I agree that the title of that question could use work. $\endgroup$ – Mason Jul 16 '18 at 23:20
  • $\begingroup$ @Mason In question 2, he asks if $C_k$ is algebraic which is not my question. My question is why or how the other ratio appears. $\endgroup$ – aleden Jul 16 '18 at 23:26
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The key is the link between the Dedekind eta function and elliptic integrals.


Let $\tau$ be purely imaginary and in upper half of complex plane and let $$q=\exp(2\pi i\tau) \in(0,1)$$ be the corresponding nome. Consider the elliptic modulus $k\in(0,1)$ corresponding to nome $q$ given in terms of $q$ via Jacobi theta functions $$k=\frac{\vartheta_{2}^{2}(q)}{\vartheta _{3}^{2}(q)},\,\vartheta_{2}(q)=\sum_{n\in\mathbb {Z}} q^{(n+(1/2))^{2}},\,\vartheta _{3}(q)=\sum_{n\in\mathbb {Z}} q^{n^2}\tag{1}$$ Let $k'=\sqrt {1-k^2}$ and we further define elliptic integrals $$K=K(k) =\int_{0}^{\pi/2}\frac{dx} {\sqrt{1-k^2\sin^2 x}}, \, K'=K(k') \tag{2}$$ The circle of these definitions is finally completed by the formula $$\frac{K'} {K} =-2i\tau\tag{3}$$ Let $\tau'$ be another purely imaginary number in upper half of the complex plane such that $$\frac{\tau'} {\tau} =r\in\mathbb {Q} ^{+} \tag{4}$$ Let the corresponding nome be $q'=\exp(2\pi i\tau') $ and the elliptic moduli be $l, l'=\sqrt{1-l^2}$ and the elliptic integrals based on these moduli be denoted by $L, L'$. Then from the relation $\tau'=r\tau$ we get via $(3)$ the modular equation $$\frac{L'} {L} =r\frac{K'} {K}, r\in\mathbb {Q} ^{+} \tag{5}$$ Under these circumstances Jacobi proved using the transformation of elliptic integrals that the relation between moduli $k, l$ is algebraic and the ratio $K/L$ is an algebraic function of $k, l $.

The Dedekind's eta function is related to elliptic integrals via the relation $$\eta(\tau) =q^{1/24}\prod_{n=1}^{\infty} (1-q^n)=2^{-1/6}\sqrt{\frac{2K}{\pi}}k^{1/12}k'^{1/3}\tag{6}$$ Now let $\tau=i/2$ so that $q=e^{-\pi} $ and then from $(3)$ we have $K=K'$ so that $k=k'=1/\sqrt{2}$ and it is well known that for this value of $k$ we have $$K(k) =\frac{\Gamma^{2}(1/4)} {4\sqrt{\pi}} \tag{7}$$ From $(6)$ it now follows that $\eta(\tau) =\eta(i/2)$ is an algebraic multiple of $\Gamma (1/4)\pi^{-3/4}$.

Let $\tau'=ri, r\in \mathbb {Q} ^{+} $ so that $\tau'/\tau=2r$ is a positive rational number. As noted above if $l, L$ correspond to $\tau'$ then the relation between $l$ and $k=1 /\sqrt{2}$ is algebraic so that $l$ is an algebraic number and the ratio $K/L$ is an algebraic function of $k, l $ and thus $K/L$ is also an algebraic number. Thus from equation $(6)$ it follows that $\eta(ri) $ is an algebraic multiple of $\Gamma (1/4)\pi^{-3/4}$.

More generally it can be proved that if $r$ is a positive rational number then the value of $\eta(i\sqrt{r}) $ can be expressed in terms of values of Gamma function at rational points and $\pi$ and certain algebraic numbers.


Also let me complete the link between $\Gamma (1/4)$ and elliptic integrals starting with your approach. We have $$\frac{\Gamma ^2(1/4)}{2\sqrt{\pi}}=\int_{0}^{\pi}\frac{dx}{\sqrt{\sin^4 x+\cos^4 x}}=\int_{0}^{\pi}\frac{dx}{\sqrt{1-2\sin^2 x\cos^2 x}}$$ and the integral can further be written as $$\int_{0}^{\pi}\frac{dx}{\sqrt{1-(1/2)\sin^2 2x}}$$ Putting $2x =t$ we can see that it reduces to $$\frac{1}{2}\int_{0}^{2\pi}\frac{dt}{\sqrt{1-(1/2)\sin^2 t}}=2\int_{0}^{\pi/2}\frac{dx}{\sqrt{1-(1/2)\sin^2 x}}=2K(1/\sqrt{2})$$ and we are done.

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  • $\begingroup$ Thank you for that in depth explanation! I was also unsure that I was heading in the right direction, but thank you for showing me where my approach would have gotten me. $\endgroup$ – aleden Jul 22 '18 at 14:48
  • $\begingroup$ forgive me if I am missing something obvious, but which trig identity did you use in the first line of completing my approach? $\endgroup$ – aleden Jul 22 '18 at 15:15
  • $\begingroup$ Never mind, I see that you use power reduction formulas. I should have attempted to reduce the integral in trig form, would have been easier than trying to do that with the non-trig form of the integral. $\endgroup$ – aleden Jul 22 '18 at 15:27
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    $\begingroup$ @aleden: from my very limited experience with elliptic integrals I have found the trigonometric form more useful. $\endgroup$ – Paramanand Singh Jul 22 '18 at 15:30
  • $\begingroup$ I can see why, due to the abundance of useful trig identities, it makes it a much easier integral to manipulate. $\endgroup$ – aleden Jul 22 '18 at 15:32
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The Dedekind eta function is related to the Euler $\phi$ function by $$ \eta(\tau)=q^{1/24}\phi(q) $$ so, for example, $$ \eta(i)=e^{-\pi/12}\phi(e^{-2\pi}). $$ In his "lost notebook", Ramanujan reported finding special values of the Euler function, such as $$ \phi(e^{-2\pi})=\frac{e^{\pi/12}\Gamma(\frac{1}{4})}{2\pi^{3/4}} $$ and therefore $$ \eta(i)=\frac{\Gamma(\frac{1}{4})}{2\pi^{3/4}}. $$ The special values that Ramanujan found have been proved by George Andrews and Bruce Berndt. See Ramanujan's lost notebook.

The Wikipedia article on the Euler function says that Ramanujan found values for $\phi(e^{-\pi})$, $\phi(e^{-2\pi})$, $\phi(e^{-4\pi})$, and $\phi(e^{-8\pi})$, which correspond to $\eta(i/2)$, $\eta(i)$, $\eta(2i)$, and $\eta(4i)$. However, as you mention, the Wikipedia article on the eta function reports as value for $\eta(3i)$ so Ramanujan must have also found $\phi(e^{-6\pi})$. I doubt that $\eta(ki)$ is known, but you have a plausible conjecture.

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    $\begingroup$ Wikipedia gives a closed form for $\eta(3i)$ and I suspect that maybe $\Gamma(\frac{1}{4})$ is connected to an elliptic integral, which perhaps implies a connection with the Dedekind Eta Function due to its relation with Jacobi Theta Functions, in which the elliptic integral can be expressed in. $\endgroup$ – aleden Jul 16 '18 at 0:37
  • $\begingroup$ If possible, could you provide a link to the portion of Ramanujan's Notebook that proves these equations? $\endgroup$ – aleden Jul 16 '18 at 0:39
  • $\begingroup$ A paper by Berndt at faculty.math.illinois.edu/~berndt/articles/aachen.pdf has two relevant comments. The first is that the result that the value for $\phi(e^{-\pi})$ "is classical" and provides references. The second is that he, together with two others, has proved that $\phi(e^{-n\pi})/\phi(e^{-\pi})$ is algebraic for every positive integer $n$, and provides a reference for that. $\endgroup$ – G. Smith Jul 16 '18 at 0:48
  • $\begingroup$ I'm new to StackExchange. Can I not put <a> in comments? $\endgroup$ – G. Smith Jul 16 '18 at 0:51
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    $\begingroup$ Sorry... the phi function in Berndt's aachen.pdf is a curly-phi function that is a kind of theta function; it isn't the Euler phi function. It's not clear that what he proved about it is relevant to your conjecture, although it seems like it might be. $\endgroup$ – G. Smith Jul 16 '18 at 5:47
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This is a long comment:

From the identity: $\eta(\frac{-1}{\tau})=\sqrt{i\tau}\eta(\tau)$ we can derive just a few more values for $\eta$ that don't appear on the wikipedia page. Taking $\tau=ki$.

$$\eta(\frac{-1}{ki})=\sqrt{-1i^2k}\eta(ki)$$

$$\eta(\frac{i}{k})=\sqrt{k}\eta(ki)$$

So now we should be able to get just a few more: Taking $k=1,2$ we get no new info but $k=3,4$ should get us closed forms for $\eta(i/3)$ and $\eta(i/4)$. So in this way we can see that if $\eta(ki)$ is an algebraic number times $\Gamma(\frac{1}{4})\pi^{-3/4}$ then so is $\eta(i/k)$. This conjecture would extend to the "Egyptian fractions."

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The value for $k = 6$ is,

$$\eta(6i) = \frac{1}{2\cdot 6^{3/8}} \left(\frac{5-\sqrt{3}}{2}-\frac{3^{3/4}}{\sqrt{2}}\right)^{1/6}\,\color{brown}{\frac{\Gamma\big(\tfrac{1}{4}\big)}{\pi^{3/4}}}$$

More generally, your observation that $\eta(k\,i)$ is a product of an algebraic number and that particular ratio (in brown) is correct. For $k>6$ and $\eta(\sqrt{-N})$, see this post: What is the exact value of $\eta(6i)$?

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    $\begingroup$ I see, is there any explanation for why that ratio appears in those Eta function values? I believe that it may be related to an elliptic integral but I do not know if thats correct. $\endgroup$ – aleden Jul 16 '18 at 17:27
  • $\begingroup$ This was answered by Niko Bagis the link that was given: math.stackexchange.com/questions/1334684/… $\endgroup$ – Mason Jul 16 '18 at 21:59
  • $\begingroup$ Right but the appearance of the ratio is explained by this answer I think ? $\endgroup$ – Mason Jul 16 '18 at 23:27

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