CW 11/18

Q.1

Is this correct or not?

k=.; 
(∫_(-a)^a)(E^(-I k x ξ))\[DifferentialD]x

k=.; 
(∫_(-a)^a)(E^(-I k x ξ))\[DifferentialD]x  + (∫_(-a)^a)(E^(-I k u ξ))\[DifferentialD]u

Q.2

Why is it wrong? where does it go wrong?

Q.3

g[x_,{a_,d_}]:=UnitBox[((x-d/2)/(2 a))] + UnitBox[((x+d/2)/(2 a))] ;
Manipulate[
Plot[g[x,{a,d}],{x,-5,5}
,Filling-> {1-> {0,Red}}, PlotRange->All]
,{a,0.25,1},{d,1.5,3}]

Q.4

Consider this argument:
The g function can be written as: g[x]= (Subscript[g, m])[x]+ (Subscript[g, p])[x]                                 (1)
where (Subscript[g, m])[x] is the negative-x part of the function: (Subscript[g, m])[x]= UnitBox[((x+d/2)/(2 a))]      (1.a)
and  (Subscript[g, p])[x] is the positive-x part of the function: (Subscript[g, p])[x]= UnitBox[((x-d/2)/(2 a))]         (1.b)
The functions are illustrated below:

a=1; d=3;
Grid[{Plot[#[[1]],{x,-5,5}
,Filling-> {1-> {0,#[[2]]}}, PlotRange->All] &  /@ Transpose[{{UnitBox[((x-d/2)/(2 a))] ,UnitBox[((x+d/2)/(2 a))] },{Red,Blue}}]}]

We want to evaluate the Fourier transform:    (∫_(-∞)^∞)g[x] (e^(i k x ξ )) dx          (2)

From Eq. (1) above:
                 (∫_(-∞)^∞)g[x] (e^(i k x ξ )) dx =(∫_(-∞)^∞)((Subscript[g, m])[x]+ (Subscript[g, p])[x])  (e^(i k x ξ )) dx 
                                      =(∫_(-∞)^∞) (Subscript[g, m])[x] (e^(i k x ξ )) dx  +(∫_(-∞)^∞) (Subscript[g, p])[x] (e^(i k x ξ )) dx           (3)

Since                          (Subscript[g, m])[x]=({{{, ({{1, -(1/2)≤((x+d/2)/(2 a))≤(1/2) }, {0, everywhere else}})}}) Or   =({{{, ({{1, -a≤ x+ (d/2) ≤a }, {0, everywhere else}})}})     (4)
define                        u≡ x+ (d/2) , then:
                                 (Subscript[g, m])[u]=({{{, ({{1, -a≤u≤a }, {0, everywhere else}})}})                                                     (5)

Then
                (∫_(-∞)^∞)(Subscript[g, m])[u] (e^(i k u ξ )) du  =(∫_(-∞)^(-a))(Subscript[g, m])[u] (e^(i k u ξ )) du   +(∫_(-a)^a)(Subscript[g, m])[u] (e^(i k u ξ )) du +(∫_a^∞)(Subscript[g, m])[u] (e^(i k u ξ )) du  
                                          = 0                                         +(∫_(-a)^a) (e^(i k u ξ )) du              +  0
And:

a=.;  k=.;  ξ= .;
(∫_(-a)^a) (E^(I k u ξ )) \[DifferentialD]u

Since we can do the same axis-translation for (Subscript[g, p])[x], this time, define v≡ x-(d/2)

a=.;  k=.;  ξ= .;
(∫_(-a)^a) (E^(I k v ξ )) \[DifferentialD]v

Thus: 
                (∫_(-∞)^∞)g[x] (e^(i k x ξ )) dx =(∫_(-∞)^∞)((Subscript[g, m])[x]+ (Subscript[g, p])[x])  (e^(i k x ξ )) dx 
                                      =((2 Sin[a k ξ])/(k ξ))  +((2 Sin[a k ξ])/(k ξ))    =4 ((Sin[a k ξ])/(k ξ))

Discuss the argument above.