Question

How do I can calculate the Fourier transform of $\dfrac{a}{{2\pi }}\dfrac{1}{{{a^2} + {x^2}}}$ ?

Answer 1

According to the given question, we have to calculate the Fourier transform of the given expression as $\dfrac{a}{{2\pi }}\dfrac{1}{{{a^2} + {x^2}}}$ .
So, first of all we have to understand about the Fourier transform of a function and its formula.
Fourier transformer: The Fourier transformer is a mathematical function that takes a time-based pattern as input and determines the overall cycle offset, rotation speed and strength for every possible cycle in the given pattern. The Fourier transform is applied to waveforms which are basically a function of time, space or some other variable. The Fourier transform decomposes a waveform into a sinusoid and thus provides another way to represent a waveform.
Formula used for Fourier transformer:
$\Rightarrow \hat f\left( \omega \right) = {f_\omega }\left[ {f\left( x \right)} \right] = \int_{ - \infty }^{ + \infty } {f\left( x \right)} {e^{i\omega x}}dx..............................(A)$ , as the Fourier transform of $f\left( x \right)$
$\Rightarrow \hat f\left( \omega \right) = {f_\omega }\left[ {f\left( x \right)} \right] = \dfrac{1}{{\sqrt {2\pi } }}\int_{ - \infty }^{ + \infty } {f\left( x \right)} {e^{ - i\omega x}}dx$
And defines inverse Fourier transform as:
$\Rightarrow f\left( x \right) = {f^{ - 1}}_x\left[ {\hat f\left( \omega \right)} \right] = \dfrac{1}{{\sqrt {2\pi } }}\int_{ - \infty }^{ + \infty } {\hat f\left( \omega \right)} {e^{i\omega x}}dx........................(B)$

Complete step-by-step solution:
Step 1: First of all we won’t worry about the $\dfrac{a}{{2\pi }}$ weight of the function or the $\dfrac{1}{{\sqrt {2\pi } }}$ weight of transform. So, first of all we have to just calculate ${I_x}$ by using the formulas (A) as mentioned in the solution hint.
$\Rightarrow {I_x} = \int_{ - \infty }^{ + \infty } {\dfrac{1}{{{a^2} + {x^2}}}} {e^{i\omega x}}dx$ , where $x \in R$
In order to compute this definite integral, consider the following complex variable function over domain C,
$\Rightarrow f\left( z \right) = \dfrac{1}{{{a^2} + {z^2}}}{e^{i\omega z}}$ and ${I_Z} = \oint_C {f\left( z \right)} dz$ , where C is the following semi-circular contour in the complex plane with radius R>0.
Step 2: we will restrict R to enclose the poles in the upper quadrants once have analysed the poles of
$f\left( z \right)$ . The denominator of the integrated is ${a^2} + {z^2}$ , and so we have simples poles when ${a^2} + {z^2}$ = 0 $\Rightarrow z = \pm ai$ .
Now, we have to be concerned with poles in ${Q_1}$ and ${Q_2}$ , that lie within our contour C that is the pole $z = ai$ , assuming a is positive.
So, then,
$\Rightarrow \oint_C {f\left( z \right)} dz = \int_{ - R}^{ + R} {f\left( x \right)dx + \int\limits_{\gamma R} {f\left( z \right)} } dz....................(1)$
Step 3: Now, we have to use residue theorem as mentioned below,
$\Rightarrow \oint_C {f\left( z \right)} dz = 2\pi i \times re{s_{z = a}}f\left( z \right)$
Now, we have to calculate residue as mentioned below,
$\Rightarrow re{s_{z = a}} = \mathop {\lim }\limits_{z \to \alpha } \left( {z - \alpha } \right)f\left( z \right)$ , where $\alpha = ai$
Step 4: Now, we have to put the value of $f\left( z \right)$ as obtained in the solution step 1 in the expression obtained in the solution step 3.
$\Rightarrow re{s_{z = a}} = \mathop {\lim }\limits_{z \to \alpha } \left( {z - \alpha } \right)\dfrac{1}{{{a^2} + {z^2}}}{e^{i\omega z}}$
Now, we have to put the value of $\alpha$ as $ai$ in the expression obtained just above.
$\Rightarrow re{s_{z = a}} = \mathop {\lim }\limits_{z \to ai} \left( {z - ai} \right)\dfrac{1}{{{a^2} + {z^2}}}{e^{i\omega z}}$
$\mathop { \Rightarrow \lim }\limits_{z \to ai} \left( {z - ai} \right)\dfrac{1}{{\left( {z - ai} \right)\left( {z + ai} \right)}}{e^{i\omega z}}$
Now, we have to put the limits after eliminating the term $\left( {z - ai} \right)$
$\Rightarrow \dfrac{1}{{\left( {ai + ai} \right)}}{e^{i\omega ai}}$
Step 5: No, we have to know that the value of ${i^2} = - 1$ , put in the expression obtained in the solution step 4.
$\Rightarrow \dfrac{1}{{2ai}}{e^{ - a\omega }}$
Thus,
$\Rightarrow \oint_C {f\left( z \right)} dz =$ $2\pi i \times \dfrac{1}{{2ai}}{e^{ - a\omega }}$
$\Rightarrow \dfrac{{\pi {e^{ - i\omega }}}}{a}$
Step 6: Now, as we often the case with contour integrals, we find that:
$\Rightarrow \mathop {\lim }\limits_{R \to \infty } \int_{\gamma R} {f\left( z \right)} dz = 0.......................(2)$
Now, we have to substitute equation (2) in equation (1),
$\Rightarrow \oint_C {f\left( z \right)} dz = \int_{ - \infty }^{ + \infty } {f\left( x \right)dx + 0}$
Now, we have to substitute the expression obtain from the solution step 5 in the expression obtain just above,
$\Rightarrow \oint_C {f\left( z \right)} dz = \dfrac{{\pi {e^{ - i\omega }}}}{a}$

Hence, the Fourier transform of $\dfrac{a}{{2\pi }}\dfrac{1}{{{a^2} + {x^2}}}$ is $\dfrac{{\pi {e^{ - i\omega }}}}{a}$.

Note: It is necessary to understand about the Fourier transform of a function and its formula as mentioned in the solution hint.
It is necessary to understand that we won’t worry about the $\dfrac{a}{{2\pi }}$ weight of the function or the $\dfrac{1}{{\sqrt {2\pi } }}$ weight of transform.