Finding the inverse Fourier transform of $\hat{f}(\omega)=2\pi e^{-|\omega-c|}$ is not as easy as I thought.
There are two ways I investigated, and it was the table-based way that was the intended way to solve this. I wanted to use the following rules of inverse transforms:
Symmetry rule:
$2\pi f(-\omega)\longrightarrow \hat{f}(t)$
where I put
$f(\omega)=e^{-i\omega T}\hat{f}$
But the problem is that shift of the variable with the constant c, in the original function, $\hat{f}(\omega)=2\pi e^{-|\omega-c|}$ .
Alternatively, I tried to find it by using the definition of the inverse Fourier transform, but got:
\begin{equation}
\int_{-\infty}^0 e^{\omega(1+it)+c}d\omega+\int_{0}^\infty e^{\omega(it-1)-c}d\omega=\lim_{B\rightarrow \infty}\frac{e^c}{1+it}+\frac{1}{it-1}(e^c-e^B)
\end{equation}
but as can be seen, that B to infinity does not complete the integral.
Any ideas?
Thanks
Best Answer
Using the definition of the inverse Fourier transform: $$\begin{split} \int_{-\infty}^{+\infty}e^{i\omega t}e^{-|\omega-c|}d\omega &= \int_{-\infty}^{c}e^{i\omega t}e^{-(c-\omega)}d\omega + \int_{c}^{+\infty}e^{i\omega t}e^{-(\omega-c)}d\omega\\ &= e^{-c}\left[\frac{e^{\omega(it+1)}}{it+1}\right]_{-\infty}^c+e^{c}\left[\frac{e^{\omega(it-1)}}{it-1}\right]_c^{+\infty}\\ &=\frac{e^{ict}}{it+1}-\frac{e^{ict}}{it-1}\\ &=\frac{2e^{ict}}{1+t^2} \end{split}$$