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Let $f(x)= \begin{cases} 3, & x \in [0,\pi]\\ 2x, & x \in(\pi,2\pi] \\ 0, & x > 2 \pi \end{cases} $

Express $f$ as a Fourier integral.

I don't know which type of integral I'm supposed to use it. Normally I'd think they mean: $$f(x) = \int_{0}^{\infty} a(u) \cos ux + b(u) \sin ux \ du$$

Where $$a(u) = \frac{1}{\pi} \int_{-\infty}^{\infty} f(x) \cos ux \ dx$$ $$b(u) = \frac{1}{\pi} \int_{-\infty}^{\infty} f(x) \sin ux \ dx$$

But since $f$ is undefined for negative $x$, how would I calculate these integrals? Could I just 'forget' the negative bits and calculate that in the bounds between $(0,\infty)$? Or should I perhaps use the even (or odd) extension of $f$ and then express it as:

$$f(x) = \sqrt{\frac{2}{\pi}}\int_{0}^{\infty} F_c (u) \cos ux \ du$$ Where $$F_c (u) = \sqrt{\frac{2}{\pi}}\int_{0}^{\infty} f (x) \cos ux \ dx \ ? $$

I would be inclined to do it the second way because I was under the impression that the first way only makes sense for $f$ defined on whole $\mathbb{R}$ ?

Spine Feast
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2 Answers2

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It looks to me the way the problem is stated (and assuming $f(x) = 0 \, \forall x < 0$), that you are asked to find

$$\int_{-\infty}^{\infty} dx \, f(x) \, e^{i k x} = 3 \int_0^{\pi} dx \, e^{i k x} + 2 \int_{\pi}^{2 \pi} dx \, x \, e^{i k x}$$

You should note that, had you been asked to find a Fourier series, the function $f$ would be periodic and not $0$ when $x > 2 \pi$.

Ron Gordon
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  • I'm 99% sure they didn't mean the complex Fourier integral. And why do you assume $f$ vanishes at negative values? If it's undefined for $x<0$, would it be correct to use either of the two formulas I gave in my question? – Spine Feast Jun 20 '13 at 13:16
  • @DepeHb: it is undefined for $x<0$. Therefore, any assumptions you want to make are fine. I just reasoned that it makes no sense to define the function $f$ as zero out to infinity if you really intended it to be periodic. Or maybe there is an error? – Ron Gordon Jun 20 '13 at 13:19
  • I was under the impression that for a Fourier integral, the function need not be periodic? – Spine Feast Jun 20 '13 at 13:22
  • @DepeHb: that's my point. – Ron Gordon Jun 20 '13 at 13:22
  • Then I'm afraid I don't understand. In the lectures we've covered the first formula (with $a(u), b(u)$) for a nonperiodic function $f: \mathbb{R} \rightarrow \mathbb{R}$. Then, for a function only defined on $[0,\infty)$ we covered a cosine/sine Fourier transform which gives rise to the second formula I provided. And in this example, the function we deal with is of the second kind - therefore my thinking to use the second formula. – Spine Feast Jun 20 '13 at 13:27
  • @DepeHb: you are overthinking this. The complex FT integral is the most general case and works everywhere that the Fourier sine & cosine integrals work. In any case, how if $f$ defined on $[0,\infty)$? If it even (use cosine), odd (use sine), or neither (use complex FT as I have laid out)? The info you supplied is insufficient to determine what to do next. – Ron Gordon Jun 20 '13 at 13:32
  • $f$ is defined at the beginning of my question. It can't be odd or even because it's undefined for negative values. I can choose an even/odd extension (for simplicity I chose even), in which case I'm looking at formula #2. – Spine Feast Jun 20 '13 at 13:45
  • @DepeHb: OK, but then state what it is you intend to do. Otherwise, I am not sure what it is you are asking. – Ron Gordon Jun 20 '13 at 13:53
  • This is a problem I had on a quiz recently. The problem just gave you $f$ (as above) and said to 'express $f$ as a Fourier integral'. I did it using formula #2, but my friends said you're supposed to use #1 here, which doesn't sit well with me, given that $f$ is undefined for negative values etc. I'm just wondering who's correct in this scenario, and if my solution is defensible should I get low marks for it. – Spine Feast Jun 20 '13 at 13:57
  • @DepeHb: the second way, as you say, assume that $f$ is even. You do not know that. You know nothing about $x<0$. So I do not know what to tell you, unless you are missing something. (Sometimes you make assumptions in class that become so routine that you forget them.) – Ron Gordon Jun 20 '13 at 14:11
  • It's even because I've created an even extension of $f$. Really I should have denoted it as $g$ or something. But then the Fourier integral expression ($g(x)=\int ...$)is valid for all arguments of this $g$ so then it is definitely true for positive arguments, where $g=f$. – Spine Feast Jun 20 '13 at 14:17
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Hint: in order to apply the Fourier series theory to your function $f$ you need to introduce an appropriate periodic extension of $f$. In other words, the given $f$ must be extended periodically to whole $\mathbb R$ before applying the formulae for the Fourier coefficients. The first task is to determine which period and which periodic extension of $f$ you need.

Avitus
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  • I'm not sure what you mean by Fourier series coefficients here, seeing as the question is about a Fourier integral. – Spine Feast Jun 20 '13 at 14:19