Let $$f(t) = \begin{cases} e^{-t}, & \text{if } t>0 \\ -e^t, & \text{if } t<0 \end{cases}$$ then compute Fourier transform of $f(t)$.
By definition we know that $$\hat{f}(\xi)=\int \limits_{-\infty}^{+\infty}f(t)e^{-2\pi i t \xi}dt=\int \limits_{-\infty}^{0}f(t)e^{-2\pi i t \xi}dt+\int \limits_{0}^{+\infty}f(t)e^{-2\pi i t \xi}dt=J+I$$ Then $$I=\int \limits_{0}^{+\infty}f(t)e^{-2\pi i t \xi}dt=\int \limits_{0}^{+\infty}e^{-t-2\pi i t \xi}dt=\int \limits_{0}^{+\infty}e^{-t(1+2\pi i \xi)}dt=\dfrac{1}{1+2\pi i \xi}$$ and $$J=\int \limits_{-\infty}^{0}f(t)e^{-2\pi i t \xi}dt=-\int \limits_{-\infty}^{0}e^{t(1-2\pi i \xi)}dt=-\dfrac{1}{1-2\pi i \xi}$$
Hence $\hat{f}(\xi)=\dfrac{4\pi i\xi}{4\pi^2 \xi^2-1}$.
Is my solution correct?
Solution using distributions
The function satisfies the distributional differential equation $f'' - f = 2\delta'.$
Taking the Fourier transform of both sides gives $$(2\pi i\xi)^2 \hat{f} - \hat{f} = 2(2\pi i\xi),$$ i.e. $$\hat{f} = \frac{4\pi i\xi}{(2\pi i\xi)^2-1} = -\frac{4\pi i\xi}{4\pi^2\xi^2+1}.$$