The inverse fourier transform of $\operatorname{sinc}\left(\omega\right)$ by definition

Question

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:

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

Answer 1

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.

Answer 2

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.