Solution To Fourier Sine Transform Problem

Obtain the relevant expressions for  I₂  if t < a and I₁    if t = a, then the expression for all results (t> a, t = a, t < a) incorporated together.

Solution:

For t < a,

I₁   = - 2 pi (e ^{iz (t-a)} / 2 z )=   -ib

= - 2 pi (e ^{iz (t-a)} / - 2ib )  =   p ( e ^{b (t-a)}  ) / b

For t =a:

e ^{iz (t-a)}  =  e ⁰

ð I₁   = 2 pi (1 / 2 z )=   2 pi (1/ 2ib) = p/ b

Then:   I₁   =  p ( e ^{-b ([t-a)]}  ) / b

Similarly: I₂   =  p ( e ^{-b ([t+a)]}  ) / b

Combining the results for: t>a, t< a, t =a:

I = ¼ Re [I₁   +   I₂ ] =

 ¼ [ e ^{-b ([t-a)]}  ) / b +  p ( e ^{-b ([t+a)]}  ) / b]

=  p /4b  ( e ^{-b ([t-a)]}  + e ^{-b ([t+a)]} )