Fourier transform spectrometers use a device called Michelson interferometer. To understand how this device works, we first consider a monochromatic radiation. This radiation strikes the beam splitter, which is constructed in such a way as to let about 50% of the incoming radiation pass, reaching the mobile mirror, while the remaining radiation is reflected towards the fixed mirror. The two beams separated in this way are reflected by the fixed mirror and by the moving mirror respectively, and return to the beam splitter which recombines the beams. The recombined beam reachs the detector. Due to the mobile mirror movement, the two reflected beams have a path difference, of the order radiation wavelenght: if the path difference is equal to an integer multiple of the wavelength, the two rays add up to form a fringe of light; if the path difference is equal to a multiple of half a wavelength, the two rays add up to form a shadow fringe.
In this way an interference pattern is obtained. The intensity $I = f(x)$ of the radiation reaching the detector is given by the formula
$$I = I_0 \left(1 + \cos 2\pi \dfrac{x}{\lambda} \right)$$
where
$I_0$ = initial radiation intensity (before been splitted)
$x$ = Moving mirror offset
$\lambda$ = wavelinght
This graph is called interferogram.To change the domain of this function from $x$ offset to $\lambda$ wavenumber a Fast Fourier Transform is needed. Having never dealt with this topic in my degree course, could you tell me if this formula is correct?
$$\int_{-\infty}^{+\infty} I_0 \left(1 + \cos 2\pi \dfrac{x}{\lambda} \right) \,d\lambda$$