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Extend $f$ evenly to $F(x)=f(|x|)$ on the whole real line, and use the Fourier transform convention $\hat{F}(k)={\int }_{\mathbb{R}}F(x)e^{-ikx}dx$. Its odd sine contribution vanishes, giving $\hat{F}(k)=2{\int }_0^{\infty }f(x)\cos (kx)dx={\hat{f}}_c(k)$ for $k>0$; the transform is also even. The Fourier inversion theorem therefore gives, for $x>0$,This is the inverse Fourier cosine transform with the factor-of-two forward convention used here.

Let the two returning monochromatic fields at the chosen detector port beHere $\nu$ is wavenumber, rather than temporal frequency. An optical-cycle average uses $⟨{\cos }^2\omega t⟩=1/2$ and $⟨\cos \omega t\cos (\omega t+\phi )⟩=\cos \phi /2$. In units absorbing the detector's field-to-power conversion,The choice of output port can reverse the cosine sign. For mutually incoherent components of the radiation spectrum, add powers rather than instantaneous fields, givingSubtracting the constant term leaves the Fourier cosine transform of the detected spectral weighting $Q(\nu )$, including throughput and detector response.

The printed prefactor requires a normalization convention. In the derivation, $a$ is the amplitude of each returning field at the detector, not an entrance amplitude with input power $a^2$. For a lossless balanced beam splitter and literal input power $P_{\mathrm{in}}$, conservation of energy instead gives $P_{\mathrm{out}}=P_{\mathrm{in}}[1+\cos \phi ]/2$, so the source formula identifies $a^2=P_{\mathrm{in}}/2$. Taking both $P_{\mathrm{in}}=a^2$ and the printed formula literally would give $P_{\mathrm{out}}=2P_{\mathrm{in}}$ at zero optical path difference, a concrete normalization contradiction.