Find the Fourier transform of the function $$f(x)=\exp(-|x|)\operatorname{sgn}(x)$$ where $$\operatorname{sgn}(x) := \begin{cases} +1 & \text{if } x \geq 0\\ -1 & \text{if } x < 0\end{cases}$$
I don't know how to convert this into a form where I can just use the standard Fourier Transforms. Is this possible and could someone show me how to solve this problem please?
Best Answer
$$\hat{f}(k)=\int e^{-2\pi i xk}e^{-|x|}\operatorname{sign}(x)dx=\int_0^\infty e^{-2\pi i xk-x}dx-\int_{-\infty}^0 e^{-2\pi i kx+x}dx=\int_0^\infty e^{-(2\pi k i+1)x}dx -\int_{-\infty}^0 e^{-(2\pi i k-1)x}dx.$$
Can you go on?
Now
$$\int_0^\infty e^{-(2\pi k i+1)x}dx=\frac{1}{2\pi k i+1}, $$
as $|e^{-(2\pi k i+1)x}|=e^{-x}$ and $x\geq 0$, and
$$\int_{-\infty}^0 e^{-(2\pi i k-1)x}dx=-\frac{1}{2\pi k i-1},$$
as $|e^{-(2\pi k i-1)x}|=e^{x}$ and $x\leq 0$. In summary
$$\hat{f}(k)=\frac{1}{2\pi k i+1}+\frac{1}{2\pi k i-1}=-\frac{4\pi ki}{4\pi^2 k^2 +1}$$