# Finding the inverse Fourier transform of $\hat{f}(\omega)=2\pi e^{-|\omega-c|}$

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:

$$$$\int_{-\infty}^0 e^{\omega(1+it)+c}d\omega+\int_{0}^\infty e^{\omega(it-1)-c}d\omega=\lim_{B\rightarrow \infty}\frac{e^c}{1+it}+\frac{1}{it-1}(e^c-e^B)$$$$

but as can be seen, that B to infinity does not complete the integral.

Any ideas?

Thanks

• I don't follow your work in evaluating those integrals - in fact it's not clear to me exactly where those integrals come from in the first place. Seems to me the natural thing is to split $\int_{-\infty}^\infty=\int_{-\infty}^c+\int_c^\infty$ and then use the fact that $|\omega-c|=\omega-c$ for $\omega>c$ and $c-\omega$ for $\omega<c$. Commented Feb 25, 2022 at 13:18
• That is what I did, I just cut it short showing the result. The boundaries you mention result by substituting the exponent by $u$ Commented Feb 25, 2022 at 16:56

Using the definition of the inverse Fourier transform: $$\begin{split} \int_{-\infty}^{+\infty}e^{i\omega t}e^{-|\omega-c|}d\omega &= \int_{-\infty}^{c}e^{i\omega t}e^{-(c-\omega)}d\omega + \int_{c}^{+\infty}e^{i\omega t}e^{-(\omega-c)}d\omega\\ &= e^{-c}\left[\frac{e^{\omega(it+1)}}{it+1}\right]_{-\infty}^c+e^{c}\left[\frac{e^{\omega(it-1)}}{it-1}\right]_c^{+\infty}\\ &=\frac{e^{ict}}{it+1}-\frac{e^{ict}}{it-1}\\ &=\frac{2e^{ict}}{1+t^2} \end{split}$$
• Great, similar to mine, only that I couldn't get rid of that term with the exponential to plus infinity. How did you get that away? $\infty \cdot t - \infty$ ? Commented Feb 25, 2022 at 19:26
• The denominator tends to $\infty$ so each fraction is $0$. Commented Feb 25, 2022 at 19:33