I have to calculate the Fourier transform of the function $f(x)=sgn(x) e^{-a |x|}, a \geq 0$. After that I have to take the limit $ a \rightarrow 0$ to calculate the Fourier transform of the sign function $f(x)=sgn(x)$. $$$$ I have found that the Fourier transform of the function $f(x)=sgn(x) e^{-a |x|}$ is: $$\widetilde{f}(k)=- \frac{2ik}{a^2+k^2}$$ Is this correct??? $$$$ Then to calculate the Fourier transform of the sign function, can I take the limit $ a \rightarrow 0$ at $\widetilde{f}(k)=- \frac{2ik}{a^2+k^2}$?? Or do I have to calculate the integral from the beginning??
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1It does depend which definition you use for the Fourier transform. But your answer to the first part does look correct. For the second part, I think you are meant to let $a\to\infty$ without recalculating the whole thing. – Stephen Montgomery-Smith Apr 13 '14 at 14:15
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@StephenMontgomery-Smith Do you mean $a \rightarrow 0$?? So is the Fourier transform of the sign function $$-\frac{2ik}{k^2}=-\frac{2i}{k}$$?? – Mary Star Apr 13 '14 at 14:20
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1Oops yes, and yes. – Stephen Montgomery-Smith Apr 13 '14 at 14:37
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@StephenMontgomery-Smith In Wolfram the result is: $$\frac{i \sqrt(\frac{2}{\pi})}{k}$$ Why is it like that?? – Mary Star Apr 13 '14 at 14:59
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1Because the definition of Fourier transform isn't unique. There are various $\sqrt{2\pi}$'s and minus signs that differ from book to book. Notice that Wolfram's is a multiplicative factor of $-1/\sqrt{2\pi}$ different from your answer. – Stephen Montgomery-Smith Apr 13 '14 at 15:08
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@StephenMontgomery-Smith Aha! Ok!!! But my result also correct?? right? – Mary Star Apr 13 '14 at 15:17
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1Yes, it is correct. – Stephen Montgomery-Smith Apr 13 '14 at 15:19
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@StephenMontgomery-Smith Great!!! Thank you very much!!! :-) – Mary Star Apr 13 '14 at 15:21