A neat evaluation of an infinite matrix? Let $M_n$ be an $n\times n$ matrix defined as
$$M_n
=\left[\frac{2i+1}{2(i+j+1)}\binom{i-1/2}i\binom{j-1/2}jx^{i+j+1}\right]_{i,j=0}^n.$$
With $I_n$ the identity matrix, consider $A_n:=I_n-M_n^2$. When I computed $\det A_n$, it looks rather "ugly". However, its infinite dimensional counterpart $\det A_{\infty}$ seems to reach a neat evaluation. To avoid issues on what it means by "determinant of infinite matrix", I simply work with the following convention: $\det(A_{\infty})=\lim_{n\rightarrow\infty}\det(A_n)$. So,

Question. Is this determinantal evaluation true?
  $$\det(A_{\infty})=\sqrt[4]{1-x^2}.$$

NOTE 1. The fractional values $\binom{i-1/2}i$ are (as usual) computed via Euler's Gamma function, $\Gamma(z)$.
NOTE 2. If it helps, we make two observations: (a) both $\det A_n$ and $\sqrt[4]{1-x^2}$ are functions of $y:=x^2$; (b) as functions of $y$, the taylor series for $\det A_n$ and $\sqrt[4]{1-y}$ agree up to degree $n$.
 A: The claimed expression for the determinant indeed holds. In fact, one may explicitly determine the eigenvalue decomposition of the infinite matrix $M_\infty$ viewed as a self-adjoint operator on a Hilbert space. Its eigenvalues are
$$ \left(\frac{2}{q_x^{p+\frac12}+q_x^{-p-\frac12}}\right)_{p=0}^\infty,\qquad q_k = e^{-\pi\frac{K\left(\sqrt{1-x^2}\right)}{K(x)}}, \tag{1}
$$ where $q_k$ is the elliptic nome of modulus $x$ and $K(x)$ the complete elliptic integral of the first kind.
Once this is known one may compute
$$
\begin{aligned}
\det(I - M_\infty^2) &= \prod_{p=0}^\infty\left(1-\left(\frac{2}{q_x^{p+\frac12}+q_x^{-p-\frac12}}\right)^2\right) \\
&=\prod_{p=0}^\infty\left(\frac{1-q_x^{2p+1}}{1+q_x^{2p+1}}\right)^2 = \frac{\theta_4(0,q_x)}{\theta_3(0,q_x)}=\sqrt[4]{1-x^2},
\end{aligned}
$$
where the last two equalities follow from standard properties of the Jacobi theta functions $\theta_i$.
To see how the eigenvalue decomposition can be obtained, it is convenient to look at the Dirichlet space $\mathcal{D}$ of complex analytic functions $f$ on the open unit disk that vanish at $0$ and that have finite norm with respect to the Dirichlet inner product
$$\langle f,g\rangle_{\mathcal{D}} = \frac{1}{\pi} \int_{|z|<1} \overline{f'(z)}g'(z) \mathrm{d}^2z = \sum_{n=1}^\infty n\,\overline{[z^n]f(z)}\,[z^n]g(z).$$
A basis is given by the monomials $(e_n:=z^n)_{n\geq 1}$, which satisfy $\langle e_n,e_m\rangle_{\mathcal{D}}=n\, \delta_{n,m}$. Let us also introduce the bounded linear operator on $\mathcal{D}$ given by
$$\mathbf{R}_x f = f \circ \psi_x, \qquad \psi_x(z) = \frac{1-\sqrt{1-x z^2}}{\sqrt{x}\,z}.$$
By power series expansion one may check that
$$ \mathbf{R}_x e_p = \sum_{\ell=p}^\infty \left(\frac{x}{4}\right)^{\ell/2} \frac{p}{\ell}\binom{\ell}{(\ell+p)/2}\, 1_{\{\ell+p\text{ even}\}} e_\ell$$
and the adjoint with respect to $\langle \cdot,\cdot\rangle_{\mathcal{D}}$ is determined by
$$ \mathbf{R}_x^\dagger e_p = \sum_{\ell=1}^p \left(\frac{x}{4}\right)^{p/2} \binom{p}{(\ell+p)/2}\, 1_{\{\ell+p\text{ even}\}} e_\ell.$$
In particular, $\mathbf{R}_x\mathbf{R}_x^\dagger$ is a self-adjoint operator that preserves the even and odd functions in $\mathcal{D}$.
Up to a factor of $2$ its matrix elements on the odd monomials are precisely $[M_\infty]_{i,j}$:
$$
\begin{aligned}
2\mathbf{R}_x\mathbf{R}_x^\dagger e_{2i+1} &= \sum_{j=0}^\infty \sum_{\ell=0}^{\min(i,j)} 2\left(\frac{x}{4}\right)^{i+j+1} \binom{2i+1}{i+\ell+1}\frac{2\ell+1}{2j+1} \binom{2j+1}{j+\ell+1} e_{2j+1}\\
&= \sum_{j=0}^\infty\frac{2i+1}{2(i+j+1)}\binom{i-1/2}i\binom{j-1/2}jx^{i+j+1}e_{2j+1}\\
&= \sum_{j=0}^\infty [M_\infty]_{i,j} e_{2j+1}.
\end{aligned}
$$
In T. Budd, Winding of simple walks on the square lattice,  arXiv:1709.04042, Section 2.1, I examined a closely related operator $\mathbf{J}_x = \mathbf{R}_x^\dagger\mathbf{R}_x$, whose matrix elements count diagonal walks on $\mathbb{Z}^2$ starting on the positive $x$-axis and hitting the $y$-axis at prescribed height. By a convenient elliptic parametrization of the disc one can determine all eigenvectors of $\mathbf{J}_x$. Since $\mathbf{R}_x$ is injective, $\mathbf{R}_x\mathbf{R}_x^\dagger$ has the same eigenvalue decomposition (after applying $\mathbf{R}_x$ to the eigenvectors of $\mathbf{J}_x$) given by Proposition 9, of which the eigenvectors with odd label $m=1,3,5,\ldots$
span the odd functions in $\mathcal{D}$. The corresponding eigenvalues are precisely the ones given above in (1).
Added (22 Jan): One may also check the formula for $\det(I\pm M_\infty)$ proposed in the post by Hucht. Using the theta function product formulas we find
$$
\det(I-M_\infty) = \prod_{p=0}^\infty \frac{(1-q_x^{p+\frac12})^2}{1+q^{2p+1}} = \left(\frac{\theta_2(0,\sqrt{q_x})}{\theta_1'(0,\sqrt{q_x})}\right)^{\frac14} \frac{\theta_4(0,\sqrt{q_x})}{\theta_3(0,q_x)^{\frac12}}.
$$
With the help of $\theta_1'(0,\sqrt{q_x}) = \theta_2(0,\sqrt{q_x})\theta_3(0,\sqrt{q_x})\theta_4(0,\sqrt{q_x})$ and
$$
\theta_3(0,\sqrt{q_x})^2 = (1+x)\theta_3(0,q_x)^2,\quad \theta_4(0,\sqrt{q_x})^2 = (1-x)\theta_3(0,q_x)^2
$$
this yields
$$
\det(I- M_\infty) = \left(\frac{(1-x)^3}{1+x}\right)^{1/8}.
$$
Similarly
$$
\det(I-M_\infty) = \prod_{p=0}^\infty \frac{(1+q_x^{p+\frac12})^2}{1+q^{2p+1}} = \left(\frac{\theta_2(0,\sqrt{q_x})}{\theta_1'(0,\sqrt{q_x})}\right)^{\frac14} \frac{\theta_3(0,\sqrt{q_x})}{\theta_3(0,q_x)^{\frac12}}= \left(\frac{(1+x)^3}{1-x}\right)^{1/8}.
$$
A: Based on your comment to my posting "Conjecture for a certain Cauchy-type determinant", I found that in addition to the formula above (I write $q$ instead of $x$, as all this is related to $q$-series, $q$-products, and Jacobi elliptic functions)
$$
\det[\mathbf I_\infty - \mathbf M_\infty(q) \, \mathbf M_\infty(q)]
=(1-q^2)^{1/4} \qquad(1)
$$
another identity holds, which is closer to my case, namely
$$
\det[\mathbf I_\infty + \mathbf M_\infty^T(q) \,\mathbf M_\infty(q)]
=(1-q^2)^{-1/4}. \qquad(2)
$$
Playing around a little further using my favorite tool Mathematica, I found the two "roots" of (1),
$$
\det[\mathbf I_\infty \pm \mathbf M_\infty(q)] = 
\left[\frac{(1 \pm q)^{3}}{1 \mp q}\right]^{1/8}.
$$
I think that the resulting identity 
$$
\det[\mathbf I_\infty + \mathbf M_\infty(q)] = 
\det[\mathbf I_\infty - \mathbf M_\infty(-q)]
$$
can be proven quite easily by analyzing the series expansions.
As $\mathbf M_\infty$ and $\mathbf M_\infty^T$ do not commute, (2) cannot easily be factored. 
However, maybe this helps a bit.
Another observation is, that my problem 325886 is an infinite product of this one, as
$$
\det[\mathbf I_\infty + \mathbf X_\infty^T(q) \,\mathbf X_\infty(q)]
= \prod_{k=1}^{\infty} 
\det[\mathbf I_\infty + \mathbf M_\infty^T(q^k) \,\mathbf M_\infty(q^k)].
$$
So they are definitely related. Obviously, all these expressions are matrix versions of $q$-product formulas. For further observations, see my updated posting 325886.
