\(\cfrac { d^{ n } }{ dt^{ n } } \left\{ t^{ n -1}f(t) \right\} \overset { Fourier }{ \longrightarrow } (-1)^{(n-1)}(i2\pi f)\cfrac { d^{ n-1 } }{ df^{ n-1 } } \left\{ F(f) \right\}\)
with light speed \(t=\cfrac{x}{c}\),
\(dt=\cfrac{1}{c}dx\)
\(c^n\cfrac{d^n}{dx^n}\left(\left(\cfrac{x}{c}\right)^{(n-1)}f(\cfrac{x}{c})\right)\overset{Fourier}{\longrightarrow }(-1)^{(n-1)}(i2\pi f)\cfrac{d^{n-1}}{df^{n-1}}\left(F(f)\right)\)
What if \(f(\cfrac{x}{c})=\cfrac{c}{x}=\cfrac{1}{t}=f\)? Then it is possible to obtain, \(F(f)\) for \(\cfrac{1}{t}\) in
\(\cfrac{1}{t}\overset{Fourier}{\longrightarrow }F(f)\)
Let \(n=1\), and \(f(\cfrac{x}{c})=\cfrac{c}{x}\),
\(c\cfrac { d }{ dx } \left( \cfrac { c } { x } \right) \overset { Fourier }{ \longrightarrow } (i2\pi f)F(f)\)
\(-\left( \cfrac { c } { x } \right)^2 \overset { Fourier }{ \longrightarrow } (i2\pi f)F(f)\)
If \(f(t)=f_t\) and \(f_t\overset{Fourier}{\longrightarrow }F(f)\) then
\(f_t^2\overset{Fourier}{\longrightarrow }F(f)\star F(f)\)
where \(\star\) denotes convolution and we differentiate \(f\) in the frequency domain from \(f_t\), the function in the time domain. So,
When \(n=1\), and \(f(\cfrac{x}{c})=(\cfrac{c}{x})^2\), \(F_1(f)=F(t)\star F(f)\)
\(c\cfrac { d }{ dx } \left( \cfrac { c } { x } \right)^2 \overset { Fourier }{ \longrightarrow } (i2\pi f)F_1(f)\)
\(-2\left( \cfrac { c } { x } \right)^3 \overset { Fourier }{ \longrightarrow } (i2\pi f)F_1(f)\)
\( (i2\pi f)\left(F(f)\star F(f)\right)=-2F(t)\star F(f)\star F(f)\) --- (2)
Consider, \(n=2\) and \(f(\cfrac{x}{c})=\left(\cfrac{c}{x}\right)^2\),
\(c^2\cfrac{d^2}{d\,x^2}\left(\cfrac{c}{x}\right)\overset{Fourier}{\longrightarrow }-(i2\pi f).\cfrac{d}{df}\left(F(f)\right)\)
\((-1)(-2).\cfrac{c^3}{x^3}\overset{Fourier}{\longrightarrow }-(i2\pi f).\cfrac{d}{df}\left(F(f)\right)\)
which is,
\(2f^3\overset{Fourier}{\longrightarrow }-(i2\pi f).\cfrac{d}{df}\left(F(f)\right)\)
Since,
\(f_t^3\overset{Fourier}{\longrightarrow }(F(f)\star F(f))\star F(f)\)
where \(\star\) denotes convolution and we differentiate \(f\) in the frequency domain from \(f_t\), function in the time domain.
\(-(i2\pi f).\cfrac{d}{df}\left(F(f)\star F(f)\right)=2F(f)\star F(f)\star F(f)\) --- (3)
\(\cfrac{d}{df}\left(F(f)\star F(f)\right)=F(f)\star F(f)\)
substitute (1) in,
\(\cfrac{d}{df}\left(i2\pi f. F(f)\right)=i2\pi f. F(f)\)
One possible solution is,
\(i2\pi f. F(f)=e^{f}\)
and \(F(f)=\cfrac{1}{i2\pi f}e^{f}\)
So,
\(\cfrac{1}{t}\overset{Fourier}{\longrightarrow }\cfrac{1}{i2\pi f}e^{f}\) --- (*)
But what does (*) mean?
A plot of y=e^x/x and y=x,
And how do check such an expression? The Duality property of Fourier tranform \(F[F(t)]=f(-f)\).
Note: The change from \(dt\) to \(dx\) via \(c.t=x\) is not necessary.
\(\cfrac{1}{t}=f\) is a function that is different from \(\cfrac{1}{\Delta t}=\cfrac{1}{T}=f\) the definition for frequency. The later is divided by an time interval \(\Delta t=T\), the period of an oscillation. \(\cfrac{1}{t}\) is the reciprocal of the passage of time with respect to some origin \(t=0\). \(\Delta t=T\) is without such an origin. \(t\) is meaningless without first defining the origin, \(\Delta t\) finds a reference point when it is applied after it has been given a value.
