Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

On Wikipedia i have come across a Fourier transformation equation in exponential form and its inverse (Wiki):

$$ \begin{split} \mathcal{F}(x) &= \int\limits^{\infty}_{-\infty}\mathcal{f}(k) \, e^{2 \pi i kx} \, \textrm{d} k\\ \mathcal{f}(k) &= \int\limits^{\infty}_{-\infty}\mathcal{F}(x) \, e^{2 \pi i kx} \, \textrm{d} x \end{split} $$

but i allso found that there is a trigonometric form of Fourier transformation (PDF, page 2)

$$ \begin{split} \mathcal{F}(x) &= \int\limits^{\infty}_{-\infty} f(k) \cos(kx) \, \textrm{d}k\\ \mathcal{f}(k) &= \int\limits^{\infty}_{-\infty} \mathcal{F}(x) \cos(kx) \, \textrm{d}x \end{split} $$


Could someone show me, how these pairs of equations are connected?


(i) I think that $\textrm{d}x$ is used for spatial integration (please correct me if i am wrong).

(ii) I think that $\textrm{d}k$ is used for integration over wave vector (please correct me if i am wrong) .

share|cite|improve this question

migrated from Feb 26 '13 at 0:05

This question came from our site for active researchers, academics and students of physics.

up vote 4 down vote accepted

Do you know about Euler's Formula?

$e^{ix} = cos(x) + isin(x)$

I think that's a hint...

Also, see this formula The first one in this column is simply the fourier integral applied over the above relation. Everything else is just a simplification into cosine

And this is even better...a whole justification of the connection

share|cite|improve this answer
Second link you provided is suplying link between time dependant and frequency dependant fourier. I have wave vector ependant and spatial dependant fourier. – 71GA Feb 25 '13 at 23:53
Call it time or frequency or wave or space --- the formulas are the same, no? – Gerry Myerson Feb 26 '13 at 0:17

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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