# Formula to Compute sigma8 for correction in non-linear regime

I try to explain my issue on this forum since it seems to be appropriate.

I need to apply a correction on $$\sigma_{8}$$ between linear and non-linear regime to keep it fixed (I make change the values of cosmological parameters at each iteration). I have to compute $$\sigma_{8}$$ from the $$P_{k}$$ (generated by CAMB code) and found the following relation (I put also the text for clarify the context) :

Part of this Klein Onderzoek is aimed at finding an estimate of the cosmological parameter $$\sigma_{8}$$ from peculiar verlocity data only. $$\sigma_{8}$$ is defined as the r.m.s. density variation when smoothed with a tophat-filter of radius of $$8 \mathrm{h}^{-1} \mathrm{Mpc} > .$$ The definition of $$\sigma_{8}$$ in formula-form is given by:

$$\sigma_{8}^{2}=\frac{1}{2 \pi^{2}} \int W_{s}^{2} k^{2} P(k) d k$$

where $$W_{s}$$ is tophat filter function in Fourier space:

$$W_{s}=\frac{3 j_{1}\left(k R_{8}\right)}{k R_{8}}$$

where $$j_{1}$$ is the first-order spherical Bessel function. The parameter $$\sigma_{8}$$ is mainly sensitive to the power spectrum in a certain range of $$k$$ -values. For large $$k,$$ the filter function will become negligible and the integral will go to zero. For small $$k,$$ the factor $$k^{2}$$ in combination with the power spectrum factor $$k^{-3}$$ will make sure that the integral is negligible. In other words, $$\sigma_{8}$$ is mostly determined by the power spectrum within the approximate range $$0.1 \leq k \leq 2 .$$ since $$\sigma_{8}$$ is only sensitive to a certain range of $$k,$$ any difference in the values of the Hubble uncertaintenty, the baryonic matter density and the total matter density will influence the found estimate.

Question 1) What numerical value have I got to take for $$R_{8}$$ in my code : for the instant, I put $$R_{8}= 8.0/0.67$$ : is this correct ?

Question 2) The other issue is, for each correction on $$A_{s}$$, that I find with this expression a value roughly around : $$\sigma_{8} = 0.8411 ........$$ instead of standard (fiducial) value $$\sigma_{8} = 0.8155 ........$$ : there is a 4 percent of difference between both values : is the expression above right ?

Could anyone tell me a good way to compute $$\sigma_{8}$$ from $$P_{k}$$ generated by CAMB-1.0.12 ?

Regards