Reading Fourier Transform Tables - Are They Symmetric?

$\begingroup$

We use the definitions $$F(k)=\mathcal{F}(f(x))=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} e^{-ikx}f(x) \ dx,$$ where the inverse is defined as $$f(x)=\mathcal{F}^{-1}(F(k))=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} e^{ikx}F(k) \ dk.$$

Consider the Fourier transform pair commonly found in tables:

Given the definition of the Fourier transform and its inverse as already defined, are these transformations symmetric? e.g. is $$\mathcal{F}\left(\frac{\sin(ax)}{x}\right)=\sqrt{\frac{\pi}{2}}H(a-k)H(a+k)$$the same as $$\mathcal{F}^{-1}\left(\frac{\sin(ak)}{k}\right)=\sqrt{\frac{\pi}{2}}H(a-x)H(a+x)?$$

Edit:

\begin{align} \mathcal{F}^{-1}_k(F(k)\cos(ckt))&=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} F(k) \cos(ckt) e^{ikx} \ dk \\ &=\frac{1}{2\sqrt{2\pi}}\int_{-\infty}^{\infty} F(k) (e^{ckti}+e^{-ckti}) e^{ikx} \ dk \\ &=\frac{1}{2\sqrt{2\pi}}\int_{-\infty}^{\infty} F(k) e^{ik(x+ct)} \ dk+\frac{1}{2\sqrt{2\pi}}\int_{-\infty}^{\infty} F(k) e^{ik(x-ct)} \ dk \\ &=\frac{1}{2}\left(f(x+ct)+f(x-ct)\right). \end{align}(recalling that $f(x)=\mathcal{F}^{-1}(F(k))$)

$\endgroup$

1 Answer

$\begingroup$

Almost: with your normalization, we have that$$\mathscr{F}\{f\}(x)=\mathscr{F}^{-1}\{f\}(-x)$$i.e.$$\mathscr{F}^2\{f\}(x)=f(-x)$$So if$$\mathscr{F}\left\{\frac{\sin(ax)}{x}\right\}(k)=\sqrt{\frac{\pi}{2}}H(a-k)H(a+k)$$Then$$\mathscr{F}^{-1}\left\{\frac{\sin(ak)}{k}\right\}(x)=\sqrt{\frac{\pi}{2}}H(a+x)H(a-x)$$

$\endgroup$ 10

Your Answer

Sign up or log in

Sign up using Google Sign up using Facebook Sign up using Email and Password Submit

Post as a guest

Post Your Answer Discard

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy