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I'm struggling on computing the fourier inverse transform of $\operatorname{sinc}\left(\omega\right)$ by definition, that is, given a fourier transform $F(\omega)$ the fourier inverse transform is defined to be:

$$f(t)=\frac{1}{2\pi}\int_{-\infty}^{\infty}F(\omega)e^{i\omega t}d\omega$$

After substitutions one gets:

$$f(t)=\frac{1}{2\pi}\int_{-\infty}^{\infty}\frac{1}{\omega}\sin(\omega)e^{i\omega t}d\omega$$

So substitution doesn't seem to help as I don't see any function and its derivative. Moreover, If I tried to do integration by parts, it is not obvious what are the parts.

I also tried to make use of the following identity: $$\sin(x)=\frac{e^{ix}-e^{-ix}}{2i}$$

and then I get:

$$f(t)=\frac{1}{2\pi}\int_{-\infty}^{\infty}\frac{e^{i\omega }-e^{-i\omega }}{2\omega i}e^{i\omega t}d\omega$$

Orel Danon
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  • Welcome to MSE. Please give context and show your own work, otherwise it is likely that you receive no answer on this site. – Andreas Nov 13 '19 at 22:22
  • If you know the definition (as an integral?), you should explicitly write this out in the body of the Question. This makes it easier for Readers to match up with notation you are familiar with, and it will make it easier for future Readers to benefit from your Question. – hardmath Nov 14 '19 at 03:44

2 Answers2

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I continue after your last step.

Note that $$\int_{-a}^{a} e^{-i\omega x }dx = \left.\frac{e^{-i\omega x }}{-i\omega}\right|_{-a}^a = \frac{e^{i\omega a }-e^{-i\omega a }}{\omega i}.$$ Using that with $a = 1$, you can continue

$$ f(t)=\frac{1}{2\pi}\int_{-\infty}^{\infty}\frac{e^{i\omega }-e^{-i\omega }}{2\omega i}e^{i\omega t}d\omega= \frac{1}{4\pi} \int_{-1}^{1}\left( \int_{-\infty}^{\infty}e^{i\omega(t-x)} d\omega\right) dx\\ = \frac{1}{2} \int_{-1}^{1}\delta(t-x) dx $$ with the (Dirac) delta-distribution $\delta(t-x)$. The result is $\frac{1}{2}$ for $-1 < t < 1$ and zero for other $t$. So your inverse Fourier transformation gives a "box" or rectangular window in time.

Andreas
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  • Nice, but essentially you have used the prior knowledge of the direct Fourier transform $\operatorname{sinc} (\omega)=\frac12\int_{-1}^1 e^{i\omega t}dt$. – user Nov 14 '19 at 11:21
  • @user Well, the situation with infinite integrals over integrands involving phase factors like $e^{-i\omega x }$ is such that convergence is typically achieved via distributions. So in one way or the other, you end up introducing such auxiliary integrations. – Andreas Nov 14 '19 at 12:22
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The following steps: $$\begin{align} 2\pi f(t)&=\int_{-\infty}^\infty\frac{\sin\omega}\omega e^{-i\omega t}d\omega\tag1\\ &=\int_{-\infty}^\infty\frac{\sin\omega}\omega \cos(\omega t) d\omega-i \int_{-\infty}^\infty\frac{\sin \omega}\omega \sin (\omega t) d\omega\tag2\\ &=\int_{-\infty}^\infty\frac{\sin\omega(1+t)+\sin\omega(1-t)}{2\omega} d\omega\tag3\\ &=\pi\frac{\operatorname{sgn}(1+t)+\operatorname{sgn}(1-t)}2\tag4\\ &=\pi\times\begin{cases} 0,&t<-1;\\ 1,&-1<t<1;\\ 0,&t>1. \end{cases}\tag5 \end{align}$$

lead to: $$ f(t)=\frac12\times\begin{cases} 0,&t<-1;\\ 1,&-1<t<1;\\ 0,&t>1. \end{cases}\tag6 $$

Explanations:

(1) Definition of the Fourier transform

(2) Euler formula and linearity of the integral

(3) The "imaginary" integral is $0$ as the integrand is an odd function of $\omega$. In the "real" integral the formula $2\sin(x)\cos(y)=\sin(x+y)+\sin(x-y)$ is applied.

(4) $\int_{-\infty}^{\infty}\frac{\sin(kx)}x dx=\pi\operatorname{sgn}(k)$ is used.

(5) Evaluation of the function.

(6) Final result.

Many proofs of the basic equality used in (4) can be found here.

user
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  • Using (4) is a comparable story: either you know it or you evaluate it. To evaluate that, you use auxiliary integrals and then you integrate over distributions. – Andreas Nov 14 '19 at 15:55
  • @Andreas I do not agree. The integral (4) in contradistinction to $\int_{-\infty}^\infty e^{ikx}dx$ converges in any reasonable sense. – user Nov 14 '19 at 16:15