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I want to find the inverse Fourier transform of the function $g(k) = e^{−a|k|}$, $a > 0, −∞ < k < ∞$. Now I proceed using $\mathcal F^{-1} [g(k)] = \int_{−∞}^∞ e^{ikx}g(k)$. I chose to split the integrals from $−∞$ to $0$ and $0$ to $∞$ because of the modulus and end up with $F^{-1} [g(k)] = \int_{0}^∞ e^{(ix-a)k} + \int_{−∞}^0 e^{(ix+a)k}$ My problem comes in evaluating $\dfrac{e^{(ix-a)k}}{ix-a}$ at $k=∞$?

Have I missed something where we know $(ix-a) < 0$ to then evaluate this at $0$? This is a course where I don't have a sheet of standard Fourier transforms and need to find answers using the integrals or the properties of a Fourier transform.

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  • $\begingroup$ $|e^{(ix-a)k}| = e^{-ak} \to 0$ since $a \gt 0$. $\endgroup$ – Chester Dec 23 '15 at 20:39
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As noted in the comments, $|e^{(ix-a)k}|=e^{-ak}$ with $a>0$, so $$\lim_{k\to\infty}\frac{e^{(ix-a)k}}{ix-a}=0$$

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