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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?

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$$\hat{f}(k)=\int e^{-2\pi i xk}e^{-|x|}\operatorname{sign}(x)dx=\int_0^\infty e^{-2\pi i xk-x}dx-\int_{-\infty}^0 e^{-2\pi i kx+x}dx=\int_0^\infty e^{-(2\pi k i+1)x}dx -\int_{-\infty}^0 e^{-(2\pi i k-1)x}dx.$$

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

$$\hat{f}(k)=\frac{1}{2\pi k i+1}+\frac{1}{2\pi k i-1}=-\frac{4\pi ki}{4\pi^2 k^2 +1}$$

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  • $\begingroup$ i still don't see how to finish this argument off $\endgroup$
    – DocMartin
    Dec 5 '13 at 13:47
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    $\begingroup$ I added some details to my answer $\endgroup$
    – Avitus
    Dec 5 '13 at 14:09
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Given that:

$$e^{-|x|} = \left\{ \begin{array}{l l} e^{x} & \text{if } x < 0 \\ e^{-x} & \text{if } x \geq 0 \end{array} \right. $$

$$sgn(x) = \left\{ \begin{array}{l l} -1 & \text{if } x < 0 \\ 1 & \text{if } x \geq 0 \end{array} \right. $$

And since $\mathcal{F}\{f(t) + g(t)\} = F(\omega) + G(\omega)$, you can calculate the Fourier transform for $t<0$ and $t\geq0$. Then, the sum is the solution.

As a hint (for $t<0$): $$f(t) = -e^x$$

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  • $\begingroup$ what are $f,F,g,G$ $\endgroup$
    – DocMartin
    Dec 5 '13 at 13:51
  • $\begingroup$ Upercase: Fourier transform. f and g are two arbitrary functions where Fourier transform of their sum is the sum of their transforms. Basically that's a property. Was the answer clear or do you want an extra detail? $\endgroup$
    – r_russo
    Dec 5 '13 at 14:25

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