# The linear growth factor of perturbations today should be 1 but I am unable to derive it

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?

$$D(a)=a$$