The definition of the linear growth factor $D(z)$ of perturbations in a cosmological setting is usually normalized to unity at redshift $z=0$. So,
$$\delta(z) = D(z)\times \delta(z=0) \tag{1}$$
where $\delta$ is the density contrast of a perturbation (I am looking at adiabatic perturbations of cold dark matter).
In the cosmology text book by Dodelson and Schmidt, there is an expression given to calculate the growth factor $D$ as a function of the cosmological scale factor $a$. When a cosmological constant and curvature is present, it is:
$$D_{+}(a)=\frac{5 \Omega_{\mathrm{m}}}{2} \frac{H(a)}{H_0} \int\limits_0^a \frac{d a^{\prime}}{\left(a^{\prime} H\left(a^{\prime}\right) / H_0\right)^3} \tag{2}$$
From some online reading, I found out that: $$\frac{H(a)}{H_0} = \sqrt{\Omega_{R, 0} a^{-4} + \Omega_{M, 0} a^{-3}+\Omega_{K, 0} a^{-2}+\Omega_{\Lambda, 0}} \tag{3}$$
In eqn (3), the omega parameters are the density ratio parameters (relative to the critical density) today. I used the following values:
$\Omega_{R, 0} = 9.23640\times 10^{-5}$, $\Omega_{M, 0} = 0.3106$, $\Omega_{\Lambda, 0} = 0.6894$ and $$\Omega_{K, 0} = 1 - \Omega_{R, 0} - \Omega_{M, 0} - \Omega_{\Lambda, 0}$$
Performing the integration in eqn (2) numerically in Python and setting $a = 1$, I get: $$D_{+}(a=1)= 0.7848$$
But I would expect the growth factor $D$ today ($z=0$ or $a=1$) to be equal to 1. I am not sure what exactly I am doing wrong or if I have incorrectly interpreted the the expression for growth factor. Can someone help?
