To $f$ and Forward

From the post "What Time?" dated 08 Jul 2016,

$\cfrac { d^{ n } }{ dt^{ n } } \left\{ t^{ n -1}f(t) \right\} \overset { Fourier }{ \longrightarrow  } (-1)^{(n-1)}(i2\pi f)\cfrac { d^{ n-1 } }{ df^{ n-1 } } \left\{ F(f) \right\}$

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,

$i2\pi f. F(f)=-F(f)\star F(f)$  --- (1)

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)

Comparing (2) and (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.