I was looking at this exercise

Evaluate the Fourier transform of the following function $$f(x)=\frac{xe^{5ix}}{x^2+2x+10}$$

and the professor was using some techniques which I don't quite understand. I think that has something to do with the properties for translation and scaling for the Fourier transform, but I cannot work it out.

First of all, we define the Fourier transform in the following manner $$\hat{f}(k)={1\over {\sqrt{2\pi}}}\int_{\mathbb{R}}e^{ikx}f(x)dx$$ Then the steps that the professor takes are as follows: $$\begin{align} &&\frac{1}{a^2+x^2}&&\overset{\mathcal{F}}{\longrightarrow} &&\frac{\pi}{a}e^{-a|k|}\tag1 \\ &[x\rightarrow x+1]&\frac{1}{x^2+2x+1+a^2}&&\overset{\mathcal{F}}{\longrightarrow} &&\frac{\pi}{a}e^{-a|k|+ik}\tag2 \\ &[a=3]&\frac{1}{x^2+2x+10}&&\overset{\mathcal{F}}{\longrightarrow} &&\frac{\pi}{3}e^{-3|k|+ik}\tag3 \\ &[f(x)\rightarrow xf(x)]&\frac{x}{x^2+2x+10}&&\overset{\mathcal{F}}{\longrightarrow} &&\frac{\pi}{3}e^{-3|k|+ik}(1+3i\operatorname{sgn}(k))\tag4\\ &[f(x)\rightarrow e^{5ix}f(x)]&\frac{xe^{5ix}}{x^2+2x+10}&&\overset{\mathcal{F}}{\longrightarrow} &&\frac{\pi}{3}e^{-3|k-5|+i(k-5)}(1+3i\operatorname{sgn}(k-5))\tag5 \end{align}$$ From what I undestrand the second step takes advantage of the translation property for the Fourier transform. The second step is trivial (clever, I know). The problems start at the last two steps: my thought was that for the fourth step he's using the convolution theorem, mainly

$$\mathcal{F}\{f\cdot g\} = \mathcal{F}\{f\}*\mathcal{F}\{g\}$$

and easily the Fourier transform of $x$ is just a Dirac delta, so the convolution is trivial. On the last step I don't have any clue on what property he's using!

Could someone explain to me what is the professor doing and if my thoughts are right or not?


For the the second-to-last step, he's using the property that F[xf] = (F[f])'/i; that is, multiplying by x and taking the Fourier Transform is the same as taking the Fourier Transform, taking the derivative of that, and then dividing by i (or, equivalently, multiplying by -i). For the last step, he's simply using the definition of the Fourier Transform: to take the Transform, you integrate $e^{ikx}f(x)$. So to get the Transform of $e^{5ix}f(x)$, you integrate $e^{ikx}e^{5ix}f(x)=e^{ikx+5ix}f(x)=e^{i(k+5)x}f(x)$. If we define $u=k+5$, then this is just the Fourier Transform of $f(x)$, except that the transformed function is a function of $u$. To get back to $k$, we subtract 5. So your professor just replaced all the instances of $k$ in the the second-to-last step with $k-5$. That is, the "k" in the second-to-last step is 5 less than the "k" in the last step.

If you're having trouble with that:


$\int e^{i(t+5)x}f(x)$

set $u = t+5$

$\int e^{iux}f(x)=F[f](u)$

since $u=t+5$, $t=u-5$, so $F[f](u)=F[f](t-5)$

So we take the Fourier Transform of $f$, then replace all instances of $k$ with $k-5$.

Remember, the Fourier Transform $F[f](k)$ is simply the strength of the $e^{ikx}$ component of $f$. By multiplying $f$ by $e^{5ix}$, we're shifting all the components by 5: the $k=0$ component becomes the $k=5$ component, etc.

There's a dual aspect to the translation property of Fourier Transforms: translating $f$ by $h$ multiplies the transform by $e^{ihx}$, and multiplying $f$ by $e^{ihx}$ translates the transform by $-h$.

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  • $\begingroup$ Precise and clear! Thank you very much $\endgroup$ – Davide Morgante Aug 2 '18 at 19:28
  • $\begingroup$ Just as an aside, is there a more general property for the Fourier transform of $xf(x)$, like some property for $$x^n f(x)$$ $\endgroup$ – Davide Morgante Aug 2 '18 at 19:30
  • 1
    $\begingroup$ @DavideMorgante Yes, I believe that the transform of $x^nf(x)$ is the nth derivative of the transform, divided by $i^n$. This is analogous to the property of Laplace Transforms that $L[x^nf]=(-1)^nL[f]^{(n)}$ where $^{(n)}$ denotes the nth derivative. $\endgroup$ – Acccumulation Aug 2 '18 at 19:49
  • $\begingroup$ You rock, thank you very much! $\endgroup$ – Davide Morgante Aug 2 '18 at 19:50

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