1
$\begingroup$

one of our homework solutions states that $$\delta(x)\equiv\frac{1}{2\pi}\int_{-\infty}^{\infty}d\omega e^{-i\omega x} $$ is the Fourier transform of the dirac delta function. But according to the definition $$\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}d\omega \delta(\omega)e^{-i\omega x}=\frac{1}{\sqrt{2\pi}}$$ must be the fourier transform. So what does that identity above mean and how was it derived?

$\endgroup$
2
$\begingroup$

Here's my best guess:

The statement in your homework solution is not giving you the Fourier transform of the dirac delta function. What it is saying is that we can define the dirac delta function as the inverse Fourier transform of $f(\omega) = 1$. That is, we can define $$ \delta(x) \equiv \frac 1{2 \pi} \int_{-\infty}^\infty d \omega \, 1 \cdot e^{i \omega x} $$ Note that it is much more typical to have the transforms be functions of $\omega$ than functions of $x$, hence my suspicions.

$\endgroup$
  • $\begingroup$ Is there a Minus missing in the Exponent? $\endgroup$ – phth Jan 23 '15 at 0:38
  • $\begingroup$ Not in the usual definition. Do you have minus there that shouldn't be there? $\endgroup$ – Omnomnomnom Jan 23 '15 at 0:40
  • $\begingroup$ In my homework solution there is one $-i\omega x$ $\endgroup$ – phth Jan 23 '15 at 0:47
  • $\begingroup$ That's strange... either way, it may be a valid definition of $\delta(x)$. It's clear they're defining $\delta(x)$ and not its Fourier transform. $\endgroup$ – Omnomnomnom Jan 23 '15 at 0:48

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.