I am trying to find the Fourier transform of $$f(x)=Ae^{-\alpha|x|}$$ where $\alpha>0$.
$f(x)$ becomes an even piecewise function defined over the intervals $-\infty$ to $0$ and $0$ to $\infty$. The corresponding figure is shown. My only question is, should I integrate over each interval separately and add the result or is there some other method? What I should get is $$F(k)= \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{0}Ae^{\alpha x}e^{-ikx}dx + \frac{1}{\sqrt{2\pi}}\int_{0}^{\infty}Ae^{-\alpha x}e^{-ikx}dx$$
Is my expression for $F(k)$ correct?
Best Answer
Your expression is correct. Further, set $x=-y$ in the first integral and observe that $$ \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{0}Ae^{\alpha x}e^{-ikx}dx = \frac{1}{\sqrt{2\pi}}\int_0^{\infty}Ae^{-\alpha y}e^{iky}dy= \left( \frac{1}{\sqrt{2\pi}}\int_{0}^{\infty}Ae^{-\alpha x}e^{-ikx}dx\right)^*, $$ where $(\cdot)^*$ denotes the complex conjugate.