What is the difference between Fourier integral and Fourier transform?

I know that for Fourier integral, the function must satisfy $\int_{-\infty}^\infty |f(t)| dt < \infty$, but what if I have a function that satisfies this condition: what does it mean to calculate Fourier transform and Fourier integral?

  • Fourier Integral: $$f(t) = \int_0^ \infty A(\omega)\cos( \omega t) + B( \omega )\sin( \omega t) d \omega$$ where $A(\omega ) = \frac{1}{\pi} \int_{- \infty }^\infty f(t)\cos( \omega t) dt$ and $B(\omega ) = \frac{1}{\pi} \int_{- \infty }^ \infty f(t) \sin( \omega t) dt$
  • Fourier Transform: $$F(\omega ) = \frac{1}{2 \pi } \int_{- \infty }^\infty f(t) e^{-i \omega t} dt$$

Fourier transform of a function $f$ is the function $\mathscr{F}f$ defined by \begin{eqnarray} \mathscr{F}f(\omega) = \frac{1}{2\pi} \int_{-\infty}^\infty f(t) e^{-i\omega t} dt \ . \end{eqnarray} Fourier integral is any integral of the form \begin{eqnarray} \int_{-\infty}^\infty y(\omega) e^{i\omega t} d\omega \ . \end{eqnarray} Fourier integral of a function $f$ is any Fourier integral, that satisfies \begin{eqnarray} x(t) = \int_{-\infty}^\infty y(\omega) e^{i \omega t} d\omega \ . \end{eqnarray} You can choose $y=\mathscr{F}x$ to find a suitable $y$.

The Fourier transform is usually defined with an expression such that it has to exist everywhere. Also the Fourier integral have to exist everywhere if we want the Fourier inversion theorem to be true. For simplicity this is usually shown using the assumption $\mathscr{F}f \in L^1$.

The Fourier integral \begin{equation} x(t) = \lim_{M \rightarrow \infty} \int_{-M}^M y(\omega) e^{i\omega t} d\omega \end{equation} has the advantage that it converges everywhere if $y=\mathscr{F}x$ and $x(t) = \mu(t) e^{-t}$, where $\mu(t)$ is the step function. However, the integral converges to the value $\frac{1}{2}(f(x+)+f(x-))$ at $0$, where $f(x+) = \lim_{t\rightarrow x^+} f(t)$ and $f(x-) = \lim_{t \rightarrow x-} f(t)$. The form has also another advantage: the residue theorem can be easily applied. However the integral is a Cauchy principal value integral and not necessarily for example an improper Riemann-integral of the first kind.

If you want to be able to transform the function $f(x) = \bigg\{ \begin{eqnarray} & \frac{\sin(x)}{x} & \ , x \neq 0 \\ & 1 & , \ x = 0 \end{eqnarray}$, you may want to write also the Fourier transform in the Cauchy principal value form. This however is not equivalent to the original definition, but an extension of it. The same result as in the extension can be also achieved using cosine and sine transforms and the definition \begin{eqnarray} \mathscr{F} = \mathscr{F}_c + i \mathscr{F}_s \ , \end{eqnarray} where \begin{eqnarray} \mathscr{F}_c f(\omega) & = & \frac{2}{\pi} \int_0^\infty P_e f(t) \cos(\omega t) d\omega \\ \mathscr{F}_s f(\omega) & = & \frac{2}{\pi} \int_0^\infty P_o f(t) \sin(\omega t) d\omega \ , \end{eqnarray} where \begin{eqnarray} P_e f(t) & = & \frac{1}{2}(f(t)+f(-t)) \\ P_o f(t) & = & \frac{1}{2}(f(t)-f(-t)) \ . \end{eqnarray}


Your question is a bit ambiguous, since you don't state what you mean by fourier integral and fourier transform.

One possible source of confusion is that, while the fourier transform is indeed a linear isometry on $L^2$, the integral $$ \int_{-\infty}^\infty f(t)e^{-i2\pi\omega t} \,dt $$ does not converge for every $f \in L^2$. It does, however converge for every $f \in L^1 \cap L^2$, and the fourier transform on the full space $L^2$ can therefore be defined as the unique extension of the transform defined by the integral on $L^1 \cap L^2$. The result is then sometimes called the Fourier-Plancherel-Transform, but sometimes also simply the fourier transform on $L^2$.

Or you could simply be referring to the difference between the integral one uses to compute the coefficients of a fourier series and the integral used to define the fourier transform (on $L^2 \cap L^2$).

  • $\begingroup$ Thank you @fgp for your reply I should have stated them sorry our lecturer told us that Fourier Integral : $f(t) = \int_0^ \infty A( \omega ) cos( \omega t) + B( \omega ) sin( \omega t) d \omega$ Where : $A(\omega ) = 1/ \pi \int_ \infty ^{- \infty } f(t) cos( \omega t) dt \\ B(\omega ) = 1/ \pi \int_ \infty ^{- \infty } f(t) sin( \omega t) dt$ While Fourier Transform : $F(\omega ) = 1/2\pi \int_ \infty ^{- \infty } f(t) e^{-i \omega w} dt$ $\endgroup$ May 24 '13 at 10:50

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.