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I've found the Fourier transform $$f(t) =\begin{cases} e^{-kt},& t \geq 0 \\ 0,& \text{otherwise}\end{cases}$$ $$e^{-kt} \leftrightharpoons \frac{1}{i\omega +k}$$ and now I'm looking to find the Fourier transform of $g(t)$ when $$g(t)=f(t)-f(-t)$$

I know that the timescaling property gives $f(-t) = F(-\omega)$ but I'm not sure how this works with exponentials. It looks like $g(t)$ will diverge, since I'm subtracting $$\int_0^\infty e^{kt}e^{-i\omega t}\ dt$$ from the original tranform. Will the piecewise somehow truncate it? Also, would like to check my understanding that dealing with the -1 time scaling would otherwise give $$ F(-\omega)=\frac{1}{k-i\omega}$$ Any help or direction would be much appreciated.

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  • $\begingroup$ I've just noticed I'm not presice regarding the divergence, I'll edit to clear that up. $\endgroup$ – William Nov 5 '18 at 0:08
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No, your integration limits are wrong. You integrate from $-\infty$ to $\infty$. But now the function is $$f(-t) =\begin{cases} e^{kt},& t \leq 0\\0,& \text{otherwise}\end{cases}$$ This means that your integration limits are from $-\infty$ to $0$

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