Fourier transform of a piecewise linear which does not vanish at infinity

Question

Good evening everyone.

I wish to take the Fourier transform of the following piecewise function:

$$ f(x) = \begin{cases} \beta & \text{if } x < -\alpha \\ \frac{-\beta}{2\alpha} (x - \alpha) & \text{if } -\alpha < x < \alpha \\ 0 & \text{if } x > \alpha \end{cases} \quad (\alpha,\,\beta> 0) $$

I tried applying the definition:

$$ \begin{split} \hat{f}(\omega) & = \int_{\Bbb R} f(x) e^{-i\omega x}\,dx \\ & = \beta \int_{-\infty}^{-\alpha}e^{-i\omega x} \, dx + \frac{-\beta}{2\alpha} \int_{-\alpha}^{\alpha} (x - \alpha) e^{-i\omega x} \, dx \\ \end{split} \tag{1} $$

But clearly, this integral does not converge. On the other hand, I read that

$$ \hat{u}(\omega) = \frac{1}{i\omega} + \pi\delta(\omega) \tag{2}$$

where $u$ is the unit step function

$$ u(x) = \begin{cases} 0 & \text{if } x < 0 \\ 1 & \text{if } x > 0 \end{cases} $$

and $\delta$ is the Dirac delta function. I don't completely understand $(2)$ but I take it for granted.

Then, I wrote $f$ as a sum of unit step functions as follows:

$$ f(x) = \beta u(-x-\alpha) + \frac{-\beta}{2\alpha} (x - \alpha)(u(x+\alpha) - u(x-\alpha)) \tag{3}$$

I think I can use $(2)$ and the linearity of the Fourier transform and complete the calculation.

Is this a valid approach? Any references would be appreciated, thank you.

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Edit: I have continued my calculation based on Jean Marie's answer (and dropped the $\beta$ term):

$$ \hat{f}(k)/\beta = \delta(k) + \frac{1}{2\alpha} \hat{r}(k) (-2i) \sin(2\pi\alpha k) $$

$$ = \delta(k) + \frac{1}{2\alpha} \left( \frac{i}{4\pi}\delta'(k)-\frac{1}{4\pi^2 k^2} \right) (-2i) \sin(2\pi\alpha k) $$

$$ = \delta(k) + \frac{1 - i\pi k^2\delta'(k)}{4\alpha\pi^2 k^2} \, i \sin(2\pi\alpha k) $$

Apply the identity $k^2\delta'(k)=0$

$$ = \delta(k) + \frac{1}{4\alpha\pi^2 k^2} \, i \sin(2\pi\alpha k) $$

Then use $\text{sinc}(x) = \sin(\pi x)/\pi x$

$$ = \delta(k) + \frac{1}{2\pi k} i \, \text{sinc}(2\alpha k) $$

Answer 1

A simpler approach than the one I had proposed in my comment.

I have taken the "other" definition of Fourier Transform (I am used to it in the framework of Signal Processing) :

$$\hat{f}(k) := \int_{\Bbb R} f(x) e^{-2 i \pi k x}\,dx.$$

Let us consider the "ramp function" : $r(x)=\max(x,0)$ whose Fourier Transform is known to be :

$$\hat{r}(k)=\dfrac{i}{4 \pi}\delta'(k)-\dfrac{1}{4 \pi^2 k^2}\tag{1}$$

(see (How does one derive the Fourier transform of the Ramp function?)).

As your function can be written

$$f(x)=\beta+\dfrac{\beta}{2\alpha}(r(x-\alpha)-r(x+\alpha)), \tag{2}$$

its Fourier Transform is easy to find using (1).

Remarks :

1) expression (2) is a cousin of your expression (3).

2) All this makes sense only in the distributional context. For example, $\dfrac{1}{4 \pi^2 k^2}$ in (1) is not integrable ; it is in fact a distribution called the Finite Part of $\dfrac{1}{4 \pi^2 k^2}.$

3) About the physical meaning of $\delta'$ as a "doublet", see the answer I gave here.