How to derive relationship between Dedekind's $\eta$ function and $\Gamma(\frac{1}{4})$ 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 $$\frac{\Gamma(\frac{1}{4})\pi^{-3/4}}{2}=\frac{\theta_3(q)}{\sqrt{2c}}$$
 A: 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.
A: 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.
A: 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."
A: 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 $\eta(6i)$?
