I am reading some paper, and I am having some trouble with some Fourier transform,

Suppose that $F(\omega)$ is the Fourier transform of $f(x)$, i.e. where

$$F(\omega)=\int_{-\infty}^{\infty}f(x)e^{-i\omega x}\,dx.$$

What is $$\frac{1}{2\pi}\int_{-\infty}^{\infty}\frac{F(\omega-a)}{i\omega}e^{i\omega x}\,d\omega ?$$ I was thinking of using the time integration property, since $i\omega$ appears at the denominator, but I am not sure how to start it.


1 Answer 1



$$f(x)=\frac{1}{2\pi}\int_{-\infty}^{\infty}F(\omega)e^{i\omega x}d\omega$$

you are looking for the inverse transform of


If you look at a table of Fourier transform correspondences you will find

$$\frac{F(\omega)}{i\omega}\Longleftrightarrow \int_{-\infty}^x f(y)dy$$


$$F(\omega-a)\Longleftrightarrow e^{iax}f(x)$$

Consequently, the function whose Fourier transform is $G(\omega)$ is the integral of a modulated version of $f(x)$:

$$G(\omega)\Longleftrightarrow \int_{-\infty}^x e^{iay}f(y)dy$$


The correspondence

$$\frac{F(\omega)}{i\omega}\Longleftrightarrow \int_{-\infty}^x f(y)dy\tag{1}$$

assumes that

$$F(0)=\int_{-\infty}^{\infty} f(x)dx=0$$

This assumption is often made implicitly because it is a standard requirement (obviously, a sufficient one) for the Fourier transform to exist. If

$$g(x)=\int_{-\infty}^x f(y)dy$$

it simply means that $g(x)$ is a well-behaved function satisfying $\lim_{x\rightarrow\infty}g(x)=0$. However, it is also possible to compute the Fourier transform of functions that do not satisfy this condition. In this case, the correspondence (1) has to be modified:

$$\frac{F(\omega)}{i\omega}+\pi F(0)\delta(\omega)\Longleftrightarrow \int_{-\infty}^x f(y)dy\tag{2}$$

which obviously simplifies to (1) if $F(0)=0$. If this condition is not satisfied, the result for the inverse transform of $G(\omega)$ given above must be modified by an additive constant:

$$G(\omega)\Longleftrightarrow \int_{-\infty}^x e^{iay}f(y)dy-\frac{1}{2}\int_{-\infty}^{\infty}e^{iax}f(x)dx$$

  • $\begingroup$ Hi, where did you find the table of correspondences? Because in some websites i found this, $$\frac{F(\omega)}{i\omega}+\pi F(0)\delta(\omega)\Longleftrightarrow \int_{-\infty}^x f(y)dy$$ instead of the one you wrote. $\endgroup$ Commented Mar 6, 2014 at 10:37
  • $\begingroup$ Yes, you can find both, depending on where you look. The one you gave is the "more correct" one of the two. However, in practice it is usually assumed that $F(0)=0$, which is equivalent to $\int_{-\infty}^{\infty}f(y)dy=0$. Then the additive term disappears. This last condition is usually taken as a basic condition for the Fourier transform of $f(x)$ to exit, in a conventional sense (i.e. without using the theory of distributions). I am sure that for the purpose of the above exercise, you can silently assume that the additive term equals zero. $\endgroup$
    – Matt L.
    Commented Mar 6, 2014 at 11:04
  • $\begingroup$ Anyway where did you find your formula? $\endgroup$ Commented Mar 6, 2014 at 11:05
  • $\begingroup$ I know it by heart ... But is given in every book on Fourier transforms. Distributions are usually introduced later on, and then you might come across the other formula. One of such books is the famous "The Fourier integral and its applications" by A. Papoulis $\endgroup$
    – Matt L.
    Commented Mar 6, 2014 at 11:11
  • $\begingroup$ What if the assumption $F(0)=0$ doesn't hold? $\endgroup$ Commented Mar 6, 2014 at 11:12

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