Find $f: \mathbb R \to \mathbb C$ such that $ \hat f (\omega) = \frac{1}{(1+i\omega)^2}. $

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If $f: \mathbb R^n \to \mathbb C$ in $L^1(\mathbb R^n)$, we define $\hat f (\xi) = \int e^{-2\pi i \langle \xi, x \rangle} f(x) dx,$ for $\xi \in \mathbb R^n$.

I'm trying to find a function $f: \mathbb R \to \mathbb C$ such that

$$ \hat f (\omega) = \frac{1}{(1+i\omega)^2}. $$

However, I have no idea how to start this problem. Any hint?

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One way is to guess, thinking about integrating by parts to get the right answer, but we can compute the inverse Fourier integral directly: $$ f(x) = \int_{-\infty}^{\infty} \frac{e^{2\pi i \omega x}}{(1+i\omega)^2} \, d\omega. $$ We can evaluate this using the Residue theorem by completing the contour in either the upper or lower half-plane with a large semicircle (and then Jordan's lemma implies the integral over the semicircle tends to zero, so it's just the sum of the residues). For $x<0$the lower half-plane contains no poles, so the integral is zero. For $x>0$, there's one double pole at $\omega = i$. One tedious computation using Laurent series later, we find the answer is $(2\pi)^2xe^{-2\pi x}$.

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Hint: Consider applying the Fourier inversion theorem.