0

I've been trying to find Fourier transform of that function.

What I got so far: - The function is even, so the FT is of order $2$, and an involution. - When we integrate the FT of that function over $\mathbb{R}$, we get $1$ (because FT of FT is the function itself, we put argument$=0~$ and $$\lim_{x\to 0} \frac{\sin (x)}{x}=1$$ well known.

But I don't know how to precisely calculate FT of that function given as a combination/mix of elementary functions. I tried also changing variables etc. yesterday but nothing worked. Thanks in advance for help.

nmasanta
  • 9,222
robin3210
  • 904
  • 4
  • 15

1 Answers1

1

Let's first define $\operatorname{sgn}t$ as $1$ if $t>0$, $-1$ if $t<0$ and $0$ if $t=0$ so (by this and this)$$\int_{\Bbb R}\frac{\sin tx}{x}dx=\pi\operatorname{sgn}t.$$

Because odd functions integrate to 0, $$\int_{\Bbb R}\frac{\sin x}{x}\exp ikxdx=\int_{\Bbb R}\frac{\sin x}{x}\cos kxdx=\frac12\int_{\Bbb R}\frac{\sin[(1+k)x]+\sin[(1-k)x]}{x}dx\\=\frac{\pi}{2}\left(\operatorname{sgn}(1+k)+\operatorname{sgn}(1-k)\right)=\pi1_{(-1,\,1)}(x).$$

J.G.
  • 115,835
  • Your first equation is not evident. – Jean Marie Sep 09 '19 at 11:27
  • @JeanMarie It's well-known; I'll see if I can edit in a link to a proof. – J.G. Sep 09 '19 at 11:31
  • 1
    Well for $t=1$ its well known, otherwise the invariance of $dx/x$ under scaling by positive numbers and the fact that sin is odd gets you the rest of the way – Calvin Khor Sep 09 '19 at 11:34
  • 1
    I've edited in a link to $t=1$ followed by a link to treatment of the sign. – J.G. Sep 09 '19 at 11:37
  • Very nice - of course I knew how to prove this, but I'd never seen such a "direct" proof. Of course it's also a little scary, since the functions are not (absolutely) integrable; the fact that if $f\in L^2$ then $\hat f(\xi)=\lim_{A\to\infty}\int_{-A}^A f(t)e^{-i\xi t}dt$ a..e. is very deep (Carleson's theorem). – David C. Ullrich Sep 09 '19 at 12:55