Sine and cosine transforms

The Fourier sine transform of  f (t), sometimes denoted by either ${\displaystyle {\hat {f}}^{s}}$ or ${\displaystyle {\mathcal {F}}_{s}(f)}$, is

${\displaystyle {\hat {f}}^{s}(\nu )=\int _{-\infty }^{\infty }f(t)\sin(2\pi \nu t)\,dt.}$

If t means time, then ν is frequency in cycles per unit time, but in the abstract, they can be any pair of variables which are dual to each other.

This transform is necessarily an odd function of frequency, i.e. for all ν:

${\displaystyle {\hat {f}}^{s}(-\nu )=-{\hat {f}}^{s}(\nu ).}$

The numerical factors in the Fourier transforms are defined uniquely only by their product. Here, in order that the Fourier inversion formula not have any numerical factor, the factor of 2 appears because the sine function has L² norm of ${\displaystyle {\tfrac {1}{\sqrt {2}}}.}$

The Fourier cosine transform of  f (t), sometimes denoted by either ${\displaystyle {\hat {f}}^{c}}$ or ${\displaystyle {\mathcal {F}}_{c}(f)}$, is

${\displaystyle {\hat {f}}^{c}(\nu )=\int _{-\infty }^{\infty }f(t)\cos(2\pi \nu t)\,dt.}$

It is necessarily an even function of frequency, i.e. for all ν:

${\displaystyle {\hat {f}}^{c}(\nu )={\hat {f}}^{c}(-\nu ).}$

Some authors[2] only define the cosine transform for even functions of t, in which case its sine transform is zero. Since cosine is also even, a simpler formula can be used,

${\displaystyle {\hat {f}}^{c}(\nu )=2\int _{0}^{\infty }f(t)\cos(2\pi \nu t)\,dt.}$

Similarly, if  f  is an odd function, then the cosine transform is zero and the sine transform can be simplified to

${\displaystyle {\hat {f}}^{s}(\nu )=2\int _{0}^{\infty }f(t)\sin(2\pi \nu t)\,dt.}$

Other authors also define the cosine transform as[3]

${\displaystyle {\hat {f}}^{c}(\nu )={\sqrt {\frac {2}{\pi }}}\int _{0}^{\infty }f(t)\cos(2\pi \nu t)\,dt.}$

and sine as

${\displaystyle {\hat {f}}^{s}(\nu )={\sqrt {\frac {2}{\pi }}}\int _{0}^{\infty }f(t)\sin(2\pi \nu t)\,dt.}$

The original function  f  can be recovered from its transform under the usual hypotheses, that  f  and both of its transforms should be absolutely integrable. For more details on the different hypotheses, see Fourier inversion theorem.

The inversion formula is[4]

${\displaystyle f(t)=\int _{-\infty }^{\infty }{\hat {f}}^{c}(\nu )\cos(2\pi \nu t)\,d\nu +\int _{-\infty }^{\infty }{\hat {f}}^{s}(\nu )\sin(2\pi \nu t)\,d\nu ,}$

which has the advantage that all quantities are real. Using the addition formula for cosine, this can be rewritten as

${\displaystyle f(t)=\int _{-\infty }^{\infty }\int _{-\infty }^{\infty }f(x)\cos(2\pi \nu (x-t))\,dx\,d\nu .}$

If the original function  f  is an even function, then the sine transform is zero; if  f  is an odd function, then the cosine transform is zero. In either case, the inversion formula simplifies.

The form of the Fourier transform used more often today is

${\displaystyle {\begin{aligned}{\hat {f}}(\nu )&=\int _{-\infty }^{\infty }f(t)e^{-2\pi i\nu t}\,dt\\&=\int _{-\infty }^{\infty }f(t)(\cos(2\pi \nu t)-i\,\sin(2\pi \nu t))\,dt&&{\text{Euler's Formula}}\\&=\left(\int _{-\infty }^{\infty }f(t)\cos(2\pi \nu t)\,dt\right)-i\left(\int _{-\infty }^{\infty }f(t)\sin(2\pi \nu t)\,dt\right)\\&={\hat {f}}^{c}(\nu )-i{\hat {f}}^{s}(\nu )\end{aligned}}}$

Using standard methods of numerical evaluation for Fourier integrals, such as Gaussian or tanh-sinh quadrature, is likely to lead to completely incorrect results, as the quadrature sum is (for most integrands of interest) highly ill-conditioned. Special numerical methods which exploit the structure of the oscillation are required, an example of which is Ooura's method for Fourier integrals[5] This method attempts to evaluate the integrand at locations which asymptotically approach the zeros of the oscillation (either the sine or cosine), quickly reducing the magnitude of positive and negative terms which are summed.

• Discrete cosine transform
• Discrete sine transform

• Whittaker, Edmund, and James Watson, A Course in Modern Analysis, Fourth Edition, Cambridge Univ. Press, 1927, pp. 189, 211