## Monday, January 9, 2023

### 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   = pi(e iz (t-a) / 2 z =   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.