# Why is this an inverse fourier cosine transform?

I would like to understand the principle of Fourier transform spectroscopy.

This is explained in Wikipedia. I did all the modeling of the system and I got the same formula:

$$I(p,\tilde{\nu}) = I(\tilde{\nu})[1 + \cos(2\pi\tilde{\nu}p)]$$

where $p$ is in $cm$ and $\tilde{\nu}$ in $cm^{-1}$. I am interested in computing $I(\tilde{\nu})$ in function of $I(p)$

$$I(p) =(p, \int_0^\infty I(p,\tilde{\nu}) d\tilde{\nu}) =(p, \int_0^\infty I(\tilde{\nu})[1 + \cos (2\pi\tilde{\nu}p)] d\tilde{\nu})$$

Up to here I'm ok with this.

But: then it is stated : "This is just a Fourier cosine transform" thus:

$$I(\tilde{\nu}) = 4 \int_0^\infty [I(p) - \tfrac{1}{2}I(p=0)] \cos (2\pi\tilde{\nu}p) dp$$

How do you conclude this? What are the missing steps in between? If the integral would have been: $$I(p) =\int_0^\infty I(\tilde{\nu}) \cos (2\pi\tilde{\nu}p) d\tilde{\nu}$$ Then Ok, it's exactly a cosine transform and I can use the inverse to compute $I(\tilde{\nu})$. But how to deal with the $[1 + \cos(...)]$?

I will soon be facing a problem where: $$I(p,\tilde{\nu}) = I(\tilde{\nu})[1 - \cos(2\pi\tilde{\nu}p)]$$ (note the sign change) and I would like to be able to perform the same computation or at least forge an intuition of what's going on here.