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?enter image description here

  • 2
    $\begingroup$ What are you doing is correct. $\endgroup$ – Mhenni Benghorbal Oct 16 '12 at 8:02
  • $\begingroup$ Perhaps you can compute these integrals? $\endgroup$ – AD. Oct 16 '12 at 14:20

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.


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