Following the method of equations (10.8)to (10.12), show that ${\mathbf{f}}_{\mathbf{1}}{\mathbf{f}}_{\mathbf{2}}$and ${\mathbf{g}}_{\mathbf{1}}\mathbf{*}{\mathbf{g}}_{\mathbf{2}}$are a pair of Fourier transforms.

Equations (10.8) to (10.12) ,

The Inverse Fourier transform of the function gis defined by the equation

$f\left(x\right)={\int }_{-\infty }^{\infty }g\left(\propto \right)e^{iax}d\propto ........\left(a\right)$

And the Fourier transform for the function f(x), is given by

$g\left(\propto \right)=\frac{1}{2\pi }{\int }_{-\infty }^{\infty }f\left(x\right)e^{-iax}dx.......\left(b\right)$

And, the convolution integral of any two function that are variable in the same variable, say$h_1\left(z\right)$and$h_2\left(z\right)$, both functions variable in z , the convolution is defined by the following

$h_1*h_2={\int }_{-\infty }^{\infty }{h}_1\left(z-x\right)h_2\left(x\right)dx$

Required to show that $g_1*g_2$and $f_1.f_2$are Fourier pair transform, which means $g_1*g_2=FT\left\{f_1.f_2\right\}$

Or, to show that the Laplace inverse of the convolution integral of the two function $g_1$and $g_2$are equal to the product of the two functions $f_1$and $f_2$

$$\begin{gathered} F{T}^{-1}\left\{g_1*g_2\right\}=F{T}^{-1}\left\{FT\left\{f_1.f_2\right\}\right\} \\ =f_1.f_2 \end{gathered}$$

Replace Z in function $h_1$with z-x , and every variable z in function $h_2$with x , such that on integration with respect to a dummy variable x , in short x is a dummy constant it can be of any symbol, but the two functions themselves are variables in z .

Thus, one must keep in mind the proper substitution for the variables for the functions in the convolution integral.

Another example for different two functions $c_1\left(y\right)$and $c_2\left(y\right)$with a different variable say y, and choose a dummy variable say k, , then

$c_1*c_2={\int }_{-\infty }^{\infty }{c}_1\left(y-k\right)c_2\left(k\right)dk$

Through this problem, the function f is the inverse of Fourier transform of the function g , and the function g is the Fourier transform of the function f .

There is many approach to such problem, It is proved that the inverse Fourier transform of the convolution of the two function is$g_1$and$g_2$are equal to the product of the two function$f_1$and$f_2$"which are defined by equation (a)", proved that both the convolution of the two functions$g_1$and$g_2$ , the product of the two functions$f_1$and$f_2$are a pair Fourier transform.

Thus, evaluate the right-hand side, of equation ($†$) thus

$f_1\left(X\right).f_2\left(X\right)={\int }_{-\infty }^{\infty }{g}_1\left(u\right)e^{ixu}du.{\int }_{-\infty }^{\infty }{g}_2\left(v\right)e^{ixv}dv$

Where, u and v are dummy constants, and hence

$={\int }_{-\infty }^{\infty }{\int }_{-\infty }^{\infty }{e}^{ix\left(u+v\right)}{g}_1\left(u\right)g_2\left(v\right)dudv$

And, let$u+v=\propto$, and hence$u=\propto -v$ , and$du=d\propto$thus

$={\int }_{-\infty }^{\infty }{\int }_{-\infty }^{\infty }{e}^{ix\left(\propto \right)}{g}_1\left(\propto -v\right)g_2\left(v\right)d\propto dv$

Change, order of integration

$={\int }_{-\infty }^{\infty }{\int }_{-\infty }^{\infty }{e}^{ix\left(\propto \right)}{g}_1\left(\propto -v\right)g_2\left(v\right)dvd\propto$

And, as $e^{ix\propto }$is constant with respect to the integration of dv , hence factor it out of the integration, hence

$={\int }_{-\infty }^{\infty }{e}^{ix\left(\propto \right)}{\int }_{-\infty }^{\infty }{g}_1\left(\propto -v\right)g_2\left(v\right)dvd\propto$

And, from the definition of the convolution,

$={\int }_{-\infty }^{\infty }{e}^{ix\left(\propto \right)}\left(g_1\left(\propto \right)*g_2\left(\propto \right)\right)d\propto$

And, from the inverse Fourier definition "equation (b)", regard the convolution as one function $g_3\left(\propto \right)$being inversely transformed to a function

$f_3\left(x\right)$, such that

$f_3\left(x\right)=f_1\left(x\right).f_2\left(x\right)$

And,

$g_3\left(\propto \right)=g_1\left(\propto \right)*g_2\left(\propto \right)$

And, hence compare equation (*) , and from the definition of the inverse Fourier transform,

$$\begin{gathered} f_3\left(x\right)={\int }_{-\infty }^{\infty }{g}_3\left(\propto \right)e^{ix\propto }d\propto \\ =F{T}^{-1}\left\{g_3\left(\propto \right)\right\} \end{gathered}$$

And hence

$f_1.f_2=F{T}^{-1}\left\{g_1\left(\propto \right)*g_2\left(\propto \right)\right\}$

And hence both the product of $f_1.f_2$and the convolution of the two functions $g_1$and $g_2$are a pair transform, also take the Fourier transform of both sides,

$$\begin{gathered} FT\left\{f_1.f_2\right\}=F{T}^{-1}\left\{g_1\left(\propto \right)*g_2\left(\propto \right)\right\} \\ {g}_1*g_2=FT\left\{f_1.f_2\right\} \end{gathered}$$