Question

Find the Fourier Transform of f(x) = (1/2a: ​

Answer 1

Answer:

this is a exam type question that is why I am not telling you the ans but i can tell you what is Fourier transform

Answer 2

Answer:

The Fourier transform for constant does not exists.

Step-by-step explanation:

Recall the definition of Fourier Transform,

Let $f(x)$ be a continuous and integrable function in $(-\infty, \infty)$ then $\hat f$ is the Fourier transform of $f(x)$ is given by

$\hat f =\frac{1}{2 \pi}\int\limits^{\infty}_{-\infty} {f(x)e^{-ipx}} \, dx$

Consider the function as follows:

$f(x)=\frac{1}{2a}$

Notice that f(x) is a constant function.

Using Fourier transform,

$\hat f =\frac{1}{2 \pi}\int\limits^{\infty}_{-\infty} {(\frac{1}{2a} )e^{-ipx}} \, dx$ (here f(x) = 1/2a)

⇒ $\hat f =\frac{1}{2 \pi} (\frac{1}{2a} )\int\limits^{\infty}_{-\infty} {e^{-ipx}} \, dx$

⇒ $\hat f =\frac{1}{4 \pi a} \int\limits^{\infty}_{-\infty} {e^{-ipx}} \, dx$ ...... (1)

Now, evaluate the integral as follows:

$\int\limits^{\infty}_{-\infty} {e^{-ipx}} \, dx =\frac{1}{-ix} (e^{-ipx})^{\infty}_{-\infty}$

                   $=\frac{1}{-ix} (e^{-ip(\infty)}-e^{-ip(-\infty)})$

                   $=\frac{1}{-ix} (e^{-\infty}-e^{\infty})$

Further, simplify as follows:

As we know $e^{-\infty}=0$ and $e^{-\infty}=\infty$.

⇒ $\int\limits^{\infty}_{-\infty} {e^{-ipx}} \, dx =\frac{1}{-ix} (0-{\infty})$

⇒ $\int\limits^{\infty}_{-\infty} {e^{-ipx}} \, dx =-\infty$

Substitute the value of integral is equation (1) as follows:

$\hat f =-\infty$

This shows that Fourier transform for constant function cannot be determined directly.

Therefore, the Fourier transform does not exists.

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