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What is the associated inverse kernel to the integral transform $T$ defined by \begin{align*} (Tf)(u) & = \int_{-\infty}^{0} \hat{f}(s)\exp((2i\pi+c)us)\ ds + \int_{0}^{+\infty} \hat{f}(s)\exp((2i\pi-c)us)\ ds\\ & = \int_{-\infty}^{+\infty} K(u,t) f(t) \ dt, \ u \in \mathbb{R^+} \end{align*} with kernel $K(u,t) = \frac{2cu}{c^2u^2+4\pi^2(u-t)^2}$, where $c > 0 $ and $\hat{f}$ is the Fourier transform of $f$. I need a closed-form of the inverse kernel $K^{−1}$.

Note that for $f(t) = \exp(2i\pi at)$, one has: \begin{align*} (Tf)(u) & = & \exp((2i\pi-c)au) \ \rm if \ \it a \rm\geq 0,\\ & & \exp((2i\pi+c)au) \ \rm if \it \ a \rm \leq 0,\\ \end{align*} that is, $T$ introduces in the usual Fourier basis an exponential decay whose rate $ca$ increases linearly with the frequency $a$.

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