# Fourier transform, time scaling with -1

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.

• I've just noticed I'm not presice regarding the divergence, I'll edit to clear that up. – William Nov 5 '18 at 0:08

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 $$0$$