4
$\begingroup$

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

$\endgroup$
  • 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
2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.