Mathematical Methods of Physics: For Undergraduate

Subsection 6.1.6 Properties of Fourier Transforms

1. Linear Property: If \(c_{1}\) and \(c_{2}\) are arbitrary constants, then

   \begin{equation*} \mathscr{F}\{c_{1}f(x) \pm c_{2}g(x)\}= c_{1}\mathscr{F}\{f(x)\} \pm c_{2}\mathscr{F}\{g(x)\} \end{equation*}

   Proof.

   we have

   \begin{equation*} \mathscr{F}\{c_{1}f(x) \pm c_{2}g(x)\}= \int\limits_{-\infty}^{\infty}e^{-isx}\left[c_{1}f(x) \pm c_{2}g(x)\right]\,dx \end{equation*}

   \begin{equation*} =\int\limits_{-\infty}^{\infty}c_{1}e^{-isx}f(x)\,dx \pm \int\limits_{-\infty}^{\infty}c_{2}e^{-isx}g(x)\,dx =c_{1}\mathscr{F}\{f(x)\} \pm c_{2}\mathscr{F}\{g(x)\} \end{equation*}
2. Change of Scale Property:
   1. If \(F(s)\) is the Fourier transform of \(f(x)\text{,}\) then \(\frac{1}{a}F\left(\frac{s}{a}\right)\) is the Fourier transform of \(f(ax)\text{.}\)

      Proof.

      We have

      \begin{equation*} \mathscr{F}\{f(x)\} =F(s) = \int\limits_{-\infty}^{\infty} e^{-isx}f(x) \,dx \end{equation*}

      \begin{equation*} \therefore \mathscr{F}\{f(ax)\}= \int\limits_{-\infty}^{\infty} e^{-isx}f(ax) \,dx = \int\limits_{-\infty}^{\infty} e^{-ist/a}f(t) \,dt/a \end{equation*}

      \begin{equation*} = \frac{1}{a}\int\limits_{-\infty}^{\infty} e^{(-is/a)t}f(t) \,dt = \frac{1}{a}F\left(\frac{s}{a}\right) \end{equation*}

      [put \(ax=t\)]
   2. \(F_{s}(s)\) is the Fourier sine transform of \(f(x)\text{,}\) then Fourier sine transform of \(f\left(\frac{x}{a}\right)\) is \(aF_{s}(as)\text{.}\)

      Proof.

      \begin{equation*} F_{s}\{f\left(\frac{x}{a}\right)\}= \int\limits_{0}^{\infty}f\left(\frac{x}{a}\right)\sin sx \,dx = \int\limits_{0}^{\infty}f(t) \sin (sat)a\,dt \end{equation*}

      \begin{equation*} =\int\limits_{0}^{\infty}af(t) \sin (sat)\,dt = aF_{s}(as) \end{equation*}

      put \(\frac{x}{a}=t\)]
   3. \begin{equation*} F_{c}\{f\left(\frac{x}{a}\right)\}= aF_{c}(as) \end{equation*}

      [We can easily proved this as above].
3. Shifting Property: If \(F(s)\) is the Fourier transform of \(f(x)\text{,}\) then \(F(s)e^{-iax}\) is the Fourier transform of \(f(x-a)\text{.}\)

   Proof.

   \begin{equation*} \mathscr{F}\{f(x-a)\} = \int\limits_{-\infty}^{\infty}f(x-a)e^{-isx} \,dx = \int\limits_{-\infty}^{\infty}f(t)e^{-is(a+t)} \,dt \end{equation*}

   put, \([x-a=t]\)

   \begin{equation*} = \int\limits_{-\infty}^{\infty} e^{-isa}.e^{-ist}f(t) \,dt \end{equation*}

   \begin{equation*} = e^{-isa}.\int\limits_{-\infty}^{\infty} e^{-ist}f(t) \,dt = e^{-isa}F(s). \end{equation*}