1

I don't know how to compute the Fourier tranform of this function:

$f(x) = \frac{\sin \pi a x}{\pi x}$

I know that $\frac{\sin \pi a x}{\pi x} = \frac{e^{i \pi a x} - e^{- i \pi a x}}{2i \pi x}$

Then I plug this function in the formula for the Fourier transform and I get:

$$\int_{\mathbb{R}}\frac{e^{i \pi a x} - e^{- i \pi a x}}{2i \pi x} e^{-i s x} dx =$$

$$= \frac{1}{2 \pi i } \int_{\mathbb{R}} \frac{e^{ix( \pi a - s)} - e^{- i x( a \pi +s)}}{x} dx=$$

$$=\frac{1}{2 \pi i } \int_{\mathbb{R}} \frac{e^{ix( \pi a - s)}}{x} dx - \int_{\mathbb{R}} \frac{e^{- i x( a \pi +s)}}{x} dx$$

What should I do know? We can get rid of $i$ in the exponent by changing variables, but that doesn't help much

Sasha
  • 363
  • 1
  • 13

1 Answers1

2

We have that: $$ g(t) = \int_{\mathbb{R}}\frac{\sin(\pi x)}{\pi x}e^{-itx}\,dx = \mathbb{1}_{(-\pi,\pi)}(t)+\frac{1}{2}\mathbb{1}_{\{\pi\}\cup\{-\pi\}}(t).\tag{1}$$

To prove such an identity, we may compute the inverse Fourier transform of $\mathbb{1}_{(-\pi,\pi)}$, or notice that:

$$ g(t) = 2\int_{0}^{+\infty}\frac{\sin(\pi x)}{\pi x}\cos(tx)\,dx = \int_{0}^{+\infty}\frac{\sin((t+\pi)x)-\sin((t-\pi)x)}{\pi x}\,dx.\tag{2}$$ However, since: $$\int_{0}^{+\infty}\frac{\sin(ax)}{x}\,dx = \frac{\pi}{2}\operatorname{sign}(a),\tag{3}$$ $(1)$ just follows. With the same steps, we have:

$$ f(t) = \int_{\mathbb{R}}\frac{\sin(\pi a x)}{\pi x}e^{-itx}\,dx = \operatorname{sign}(a)\cdot g\left(\frac{t}{|a|}\right).\tag{4}$$

Jack D'Aurizio
  • 353,855
  • I suppose we need to change variables $\pi x \rightarrow \pi a x $. However, I do not see why the formula is true. Could you add some calculations? – Sasha Dec 11 '14 at 16:35
  • @Sasha: ok, I am going to add some details. – Jack D'Aurizio Dec 11 '14 at 16:37
  • 1
    @Sasha: done. Do you need also a proof of $(3)$? It is not hard to find the proof of it on MSE. – Jack D'Aurizio Dec 11 '14 at 16:54
  • Yes, I do need a proof of this equality. Could you tell me how to prove that the limit is $\frac{\pi}{2} sign(a)$? I haven't been able to find it on MSE, although I'm sure it must be proven somewhere. – Sasha Dec 11 '14 at 17:30
  • 1
    @Sasha: with a change of variable, you only need to prove that $$\int_{\mathbb{R}}\frac{\sin x}{x},dx = \pi, $$ that is also known as the Dirichlet integral. There are really many proofs of this identity. – Jack D'Aurizio Dec 11 '14 at 17:32
  • Ok, I think I can handle it from here. Thank you! – Sasha Dec 11 '14 at 17:36