Question

find Fourier transform of f(x)=e^-|x| , -infinity to +infinity
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Answer 1

Answer:

\huge\mathcal\pink{Dear \: user}Dearuser

\huge\texttt\green{please check the question again}please check the question again

Answer 2

Given:

$f(x) = e^{-\left | x \right |}$

To find:

Fourier transform

Solution:

we know that,

     $\hat{f}(\xi ) = \int_{-\infty }^{\infty } e^{i\xi t}e^{-\left | t \right |} dt$

now,

 $\hat{f}(\xi ) = \int_{0 }^{\infty } e^{i\xi t}e^{-t} dt + \int_{-\infty }^{0} e^{i\xi t}e^t dt$

         $= \left [ \tfrac{1}{i\xi -1}.e^{(i\xi -1)t} \right]^\infty _0 + \left [ \tfrac{1}{i\xi +1}e^{(i\xi +1)t} \right ]^0_{-\infty }$

         $= \lim_{a\rightarrow \infty }\left [ \tfrac{1}{i\xi -1}.e^{(i\xi -1)t} \right]^a _0 + \lim_{a\rightarrow \infty }\left [ \tfrac{1}{i\xi +1}e^{(i\xi +1)t} \right ]^0_{-a }$

since, $\lim_{a\rightarrow \infty }e^{(i\xi -1)a}=0$  

        similarly, $\lim_{a\rightarrow \infty }e^{(i\xi +1)(-a)}=0$

so, we have

         $= -\frac{1}{i\xi -1} + \frac{1}{i\xi +1}$

thus,

         $\hat{f}{(\xi)} = \frac{2}{1+\xi ^2}$

This is the required Fourier transform.