Simplifying the Fourier transform

Perhaps the best-known application of negative frequency is the formula:

${\displaystyle {\hat {f}}(\omega )=\int _{-\infty }^{\infty }f(t)e^{-i\omega t}dt,}$

which is a measure of the energy in function ${\displaystyle f(t)}$ at frequency ${\displaystyle \omega .}$ When evaluated for a continuum of argument ${\displaystyle \omega ,}$ the result is called the Fourier transform.^{[upper-alpha 1]}

For instance, consider the function:

${\displaystyle f(t)=A_{1}e^{i\omega _{1}t}+A_{2}e^{i\omega _{2}t},\ \forall \ t\in \mathbb {R} ,\ \omega _{1}>0,\ \omega _{2}>0.}$

And:

${\displaystyle {\begin{aligned}{\hat {f}}(\omega )&=\int _{-\infty }^{\infty }[A_{1}e^{i\omega _{1}t}+A_{2}e^{i\omega _{2}t}]e^{-i\omega t}dt\\&=\int _{-\infty }^{\infty }A_{1}e^{i\omega _{1}t}e^{-i\omega t}dt+\int _{-\infty }^{\infty }A_{2}e^{i\omega _{2}t}e^{-i\omega t}dt\\&=\int _{-\infty }^{\infty }A_{1}e^{i(\omega _{1}-\omega )t}dt+\int _{-\infty }^{\infty }A_{2}e^{i(\omega _{2}-\omega )t}dt\end{aligned}}}$

Note that although most functions do not comprise infinite duration sinusoids, that idealization is a common simplification to facilitate understanding.

Looking at the first term of this result, when ${\displaystyle \omega =\omega _{1},}$ the negative frequency ${\displaystyle -\omega _{1}}$ cancels the positive frequency, leaving just the constant coefficient ${\displaystyle A_{1}}$ (because ${\displaystyle e^{i0t}=e^{0}=1}$), which causes the infinite integral to diverge. At other values of ${\displaystyle \omega }$ the residual oscillations cause the integral to converge to zero. This idealized Fourier transform is usually written as:

${\displaystyle {\hat {f}}(\omega )=2\pi A_{1}\delta (\omega -\omega _{1})+2\pi A_{2}\delta (\omega -\omega _{2}).}$

For realistic durations, the divergences and convergences are less extreme, and smaller non-zero convergences (spectral leakage) appear at many other frequencies, but the concept of negative frequency still applies. Fourier's original formulation (the sine transform and the cosine transform) requires an integral for the cosine and another for the sine. And the resultant trigonometric expressions are often less tractable than complex exponential expressions. (see Analytic signal, Euler's formula § Relationship to trigonometry, and Phasor)