Age of a dust universe without a cosmological constant ($H_0{t}_0={\int }_0^1{\left(1-2q_0+2q_0/x\right)}^{-1/2}dx$)

For an expanding pressureless matter universe with zero cosmological constant, normalize $x=a/a_0$. Dust relation between deceleration and density and mass conservation give ${\dot{x}}^2=H_0^2(1-2q_0+2q_0/x)$. Integrating from the Big Bang yields the age integral above.

For $q_0>1/2$, set $x=q_0(1-\cos \theta )/(2q_0-1)$ to obtain $H_0{t}_0=q_0({\theta }_{\ast }-\sin {\theta }_{\ast })/(2q_0-1{)}^{3/2}$, with $\cos {\theta }_{\ast }=(1-q_0)/q_0$ on the expanding branch. For $0<q_0<1/2$, the corresponding hyperbolic expression is $q_0(\sinh {\eta }_{\ast }-{\eta }_{\ast })/(1-2q_0{)}^{3/2}$ with $\cosh {\eta }_{\ast }=(1-q_0)/q_0$. The flat-dust and empty Milne universe limits are respectively $2/3$ and $1$.

For an expanding pressureless matter universe with zero cosmological constant, fixed current Hubble parameter $H_0$ and current matter cosmological density parameter $\Omega >0$, the age of a dust universe without a cosmological constant is $H_0{t}_0={\int }_0^1\sqrt{x}/\sqrt{\Omega +(1-\Omega )x}dx$. For every $0<x<1$ the integrand decreases strictly as $\Omega$ increases. Thus open models with $0<\Omega <1$ are older than the flat model, and closed expanding models with $\Omega >1$ are younger. The flat value is $2/3$ and the empty Milne universe limit is $1$. Specifying what is held fixed matters: at fixed present scale factor and energy density, the integral $t_0={\int }_0^{a_0}da/\sqrt{C/a-k}$ instead increases with $k$, provided all three models reach $a_0$ on their expanding branches.

For an expanding pressureless matter universe with positive curvature, zero cosmological constant and present scale factor one, set $u_0=\arccos (2/{\Omega }_0-1)\in (0,\pi )$. The closed matter-dominated Friedmann solution then gives the displayed age of an FLRW universe. Its flat limit is $2/(3H_0)$, despite the individually vanishing numerator and denominator. To see this, write $\delta ={\Omega }_0-1$; then $u_0=2\arctan \sqrt{\delta }$ and $u_0-\sin u_0=(4/3){\delta }^{3/2}+O({\delta }^{5/2})$.