TITLE: Age of Universe from Friedmann Equation - How to actually solve the integral?
QUESTION [7 upvotes]: The Friedmann equation for a flat universe can be written as
$$
H(t)=\frac{\dot{a}}{a}=H_0\sqrt{\Omega_{m,0}\cdot a^{-3}+\Omega_{\Lambda,0}}=H(a)
$$
To calculate the age of the universe, many books jump directly to the result. But there should be some sort of integral in between. I assume one can do the following:
$$
t=\int_0^{t_0}\!\mathrm{d}t=\int_0^{1}\!\mathrm{d}a\frac{\mathrm{d}t}{\mathrm{d}a}=\int_0^{1}\!\frac{\mathrm{d}a}{\dot{a}}
$$
with $\dot{a}$ from above expressin for $H$.
But how is this integral solved? Mathematica did something for hours but did not came up with a result. Most books and wikipedia pages skip directly to the result
$$
t_0=\frac{1}{3H_0\sqrt{\Omega_\Lambda}}\log{\frac{1+\sqrt{\Omega_\Lambda}}{1-\sqrt{\Omega_\Lambda}}}
$$
which leads to the well known result of ~13 billion years (depending on the DM density).
Again: But how is the integral solved?

REPLY [1 votes]: for flat universe ($\Omega_m + \Omega_{\Lambda}=1)$
$$H^2 = H_0^2(\Omega_ma^{-3}+\Omega_{\Lambda})$$ 
or $$\frac{\dot{a}^2}{a^2} = H_0(\Omega_ma^{-3}+\Omega_{\Lambda})$$
which becomes
$$\dot{a}^2 = H_0(\Omega_ma^{-1}+\Omega_{\Lambda}a^2)$$
taking square root
$$\dot{a} = H_0\sqrt{\Omega_ma^{-1}+\Omega_{\Lambda}a^2}$$
let us write in the form of
$$\frac{da}{dt} = H_0\sqrt{\frac{\Omega_m}{a}+\Omega_{\Lambda}a^2}$$
$$\frac{da}{\sqrt{\frac{\Omega_m}{a}+\Omega_{\Lambda}a^2}} = H_0dt$$
$$\frac{da}{\sqrt{\Omega_{\Lambda}}\sqrt{(\frac{\Omega_m}{\Omega_{\Lambda}})(\frac{1}{a}) + a^2}}= H_0dt$$
set $$\frac{\Omega_m}{\Omega_{\Lambda}} = L = 0.44927$$
(I took $\Omega_{\Lambda} = 0.69$, $\Omega_m=0.31$)
By taking $a(t_{now})=1$ 
$$\int_0^{a(t_{now})=1}\frac{da}{\sqrt{\frac{L}{a}+a^2}} = \int_0^{t_{universe}} \sqrt{\Omega_{\Lambda}}H_0dt$$
by using wolfram the solution becomes, 
$$\left.\frac{2}{3}log(a^{3/2} + \sqrt{a^2+L})\right|_0^1 = \sqrt{\Omega_{\Lambda}}H_0t_{uni}$$
or you can use https://www.integral-calculator.com to make a numerical calculation of the integral. In any case we have 
$$\int_0^1\frac{da}{\sqrt{\frac{L}{a}+a^2}} = 0.793513$$
Hence,
$$t_{uni} = \frac{0.793513}{0.83066 \times H_0}$$
For $H_0 = 68km/s/Mpc$
$$t_{uni} = 0.9552801 \times H_0^{-1}= 0.9552801 \times 14.39~\text{Gyr} = 13.74 ~\text{Gyr}$$
{"set_name": "stack_exchange", "score": 7, "question_id": 532828}