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 $$\frac{\Gamma(\frac{1}{4})\pi^{-3/4}}{2}=\frac{\theta_3(q)}{\sqrt{2c}}$$