Using The Fourier Sine Transform In Contour Integration

 Given:

I = ò^¥_o    sin tx sin ax dx/ (x² +  b ²)   a > 0,   b  > 0 

We have the actual integral for the Fourier sine transform

Or we can write: 

I = ½  ò^¥_o  sin tx sin ax dx/ (x² +  b ²)   (b  > 0 ) 

We then apply the identity:

sin tx sin ax =  ½ [cos x(t – a) – cos x(t + a)]

= ½ Re [ e ^ix(t-a)    - e ^ix(t+a) ]

Thence:

I =  ¼ Re ò^¥_-¥ [ e ^ix(t-a)    - e ^ix(t+a) ]dx / (x² +  b ²)  

=  ¼ Re [I₁   -   I₂ ]

Where:

I₁   = ò^¥_-¥   e ^ix(t-a) dx / (x² +  b ²)  

I₂   = ò^¥_-¥   e ^ix(t+a) dx / (x² +  b ²)  

Suppose t > a,   then: e ^imx  =>  m=  (t – a)

So m >0  yields the contour in upper half plane, i.e.

Then singular points occur at  + ib, but we use only ib in y > a

Hence:

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

=  p/b  ( e ^{-b (t-a)}  )  if t >a

Suggested 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.