Inverse Fourier Transform of $e^{-ck}/k$ How do I find the inverse fourier transform of a function of the form $$\hat{f}(k)=\frac{e^{-ck}}{k},$$ with $c$ being some constant (can be complex)? The definition of the inverse fourier transform that I am using is $$f(x) = \frac{1}{2\pi} \int_{-\infty}^{\infty} e^{ikx}\hat{f}(k)\, dk$$
I have tried direct integration which has led nowhere, and I cannot come up with some function which gives this as its Fourier transform.
Thanks :)
 A: If $c$ is a pure imaginary number, then the transform can be computed in the sense of tempered distributions. Let $1/k$ be the distribution defined by
$$\left( \frac 1 k, \phi \right) =
\operatorname{v.\!p.} \int_{-\infty}^\infty \frac {\phi(k)} k dk,$$
then
$$\mathcal {F} \!\left[ \frac 1 k \right] =
\left( \frac 1 k, \frac 1 {2 \pi} e^{i x k} \right) =
\frac i 2 \operatorname{sgn}x, \\
\mathcal {F} \!\left[ \frac {e^{-c k}} k \right] =
\mathcal {F} \!\left[ \frac 1 k \right](x + i c) = 
\frac i 2 \operatorname{sgn} (x + i c).$$
If $c$ has a real component, the answer will depends on the chosen space of test functions. Gelfand and Shilov's book defines the Fourier transform for distributions acting on functions with finite support. Then $\mathcal {F}[e^k]$ is well-defined, but it is a distribution living in a different dual space (acting on entire functions with a certain condition on their rate of growth). For the Schwartz space of functions of rapid decay, $e^k$ is not a valid functional, because the integral of $e^k \phi(k)$ does not necessarily converge.
A: As in @JackD'Aurizio 's observations, there was probably a prior computational error: in solving $(\Delta-s^2) u=\delta$, taking Fourier transforms gives (up to constants) $(x^2+s^2)\widehat{u}=1$, and $\widehat{u}(x)=1/(x^2+s^2)$. The inverse Fourier transform can be evaluated by residues (unsurprisingly), and/but this technique gives slightly different results depending on whether $x$ is positive or negative...
For that matter, we should certainly expect that a Fourier transform of a tempered distribution is a tempered distribution, and something like $e^x$ is definitely not a tempered distribution, whatever virtues it may have. So any computation that produces this as an alleged FT of a tempered distribution is ... doubtful.
And, then, indeed, something like $e^{-|x|}$ (with constants) is surely the right thing...
