kill(all)$

load(lrats)$

complex :proof : jordan's lemma (inverse laplace transform = 'ilt'),version 1
peter.vlasschaert@gmail.com, 06/09/2019

Bromwich contour

build with : geogebra 5 without programming. (figure1)

The inverse Laplace transform.
========================

1e) given: formula inverse laplace transform.
-------------------------------------------------------------

=>  closed contour ( C ),see figure.

p1:f(t) = 1/(2*π*%i)*'integrate(F(s)*exp(s*t),s,γ-%i*∞,γ+%i*∞);

p1a:q1=2*π*%i;

p1b:1/q1*'integrate(F(s)*exp(s*t),s,γ-%i*∞,γ+%i*∞);

2e) closed contour (=C)
==================
first integral ( = LHS , closed contour integral)
C = L+C[R]

p2a:C=integrate(F[C](s)*exp(s*t),s);
p2b:L=integrate(F[L](s)*exp(s*t),s);
p2c:C[R]=integrate(F[CR](s)*exp(s*t),s);

result
--------

p2:rhs(p2a)=rhs(p2b+p2c);

p2a:rhs(p2b) = p1b*q1;

p2b:lhs(p2) = rhs(p2a)+rhs(p2c);

3e)  jordan's lemma (proof)
=====================
C[R] when  R → ∞ : integrate(F[CR](s)*%e^(s*t),s)  → 0

p3:'limit(rhs(p2c),R,∞) = 0;

proof 'p3'
-------------

a)  assume t > 0 : CR => semicircular contour C[R]
---------------------------------------------------------------------
*   points : on the contour

p3a:s=γ+R*exp(%i*θ);    /*π/2≤θ≤ 3*π/2 */

b) F(s) → 0 as |s| → ∞
--------------------------------
*  C[R]  as R → ∞,F(s) → ∞

meaning : ε > 0, R : found such that  |F(s)| =|F(R*%e^(%i*θ)+γ)| < ε

assume(ε>0);

p3b:abs(rhs(p2c))<ε*'integrate(absc(exp(s*t)),s);

c) find absc(%e^(s*t))
------------------------------

assume(t>0,R>0);

p3c1:%e^(s*t);

p3c2:lratsubst([p3a],p3c1);

p3c3:abs(p3c2);

rem :

euler identity
==========

p3c41:exp(%i*θ) = rectform(exp(%i*θ) );

p3c42:"abs(exp(%i*θ)) " = abs(exp(%i*θ));

d) use input :p3c3 into p3b
-------------------------------------

p4:rhs(p3b);

p4a:args(p4);

p4b:args(p4a[1]);

p4c:p4b[1] = p3c3;

result :

p4d1:absc(%e^(s*t))=v(s);

how to change variable :   s to θ,see 'part e'

p4d11:v(θ) = p3c3;

p4d2:lratsubst([p4d1],p3b);

p4d2a:rhs(p4d2);

rem : v(s) = abs(v(s))
          v(θ) =abs(v(θ))

e) change integration s to θ
--------------------------------------
math : ds = i*R*exp(i*θ)*dθ

p5:changevar(%, p3a, θ, s);

p6:v(R*%e^(%i*θ)+γ) =rhs(p4c);

p6a:lratsubst([p6],p5);

intermezzo: begin

p6b1:args(p6a);

p6b2:abs(p6b1[1])*p6b1[2]*p6b1[3];

p6b3:args(p6b1[4]);

p6b4:abs(p6b3[1]);

end :intermezzo

=>  result :
from figure: < π/2≤θ≤ 3*π/2 >

assume(γ>0);

p7:p6b2*integrate(p6b4,p6b3[2],π/2, 3*π/2); 

p7a:args(p7);

p7b:args(p7a[3]);

f) interchange range  0 to π
--------------------------------------
φ = θ -π/2

p8:φ = θ -π/2;

lower part integration

p8a:lratsubst([θ=p7b[3]],p8);

upper part integration

p8b:lratsubst([θ=p7b[4]],p8);

p8c:p6b2*changevar(p7a[3], p8, φ, θ);

rem:

 ==========================================================

p8d:factor((R*t*cos((2*φ+%pi)/2)+t*γ)=trigreduce(R*t*cos((2*φ+%pi)/2)+t*γ));

 ===========================================================

result :

p8e:p6b2*integrate(exp(rhs(p8d)),φ,rhs(p8a),rhs(p8b));

f) interchange range  0 to π/2
--------------------------------------
reason : sin(α) =sin(π-α), symmetric

p9:sin(α) =sin(π-α);

rem :α=0,π

p9a:subst([α =0],p9);
p9b:subst([α =π],p9);

rem :α=0,π/2

p9aa:subst([α =0],p9);
p9bb:subst([α =π/2],p9);

plot :sin(θ)
========

wxplot2d([sin(θ),2/π*θ], [θ,0,π])$

how to find : redl ine
===============

p10a:[[0,0],[π/2,1]];

first point (x1,y1) : line
-----------------------------

p10a1:x1=(p10a[1])[1];
p10a2:y1=(p10a[1])[2];

second point (x1,y1) : line
-----------------------------------

p10b1:x2=(p10a[2])[1];
p10b2:y2=(p10a[2])[2];

p10b:float(p10a);

p10c:[p10a1,p10a2,p10b1,p10b2,x=θ];

general equation : red line

p10d:y = y1+(y2-y1)/(x2-x1)*(x-x1);

equation : red line    
==============

p10e:lratsubst(p10c,p10d);

change : lower limit to 0 and upper limit to π/2
---------------------------------------------------------------

→  0 to π       ,see plot
-------------------------------

p11:p8e;

p11a:args(p11);

p11b:args(p11a[3]);

φ back to θ  
----------------

p11c:lratsubst([φ = θ],p11b);

→  0 to π/2  ( *2), see plot:same area : 0 to π
----------------------------------------------------------------

p12:2*p6b2*'integrate(p11c[1],θ,0,π/2);

use inequality: sin(θ)≥(2*θ)/%pi

p12a:sin(θ) = rhs(p10e);

final result :integral for the proof
=========================

p13:lratsubst([p12a],p12);

p13a:ev(p13,integrate);

p13b:ratsimp(p13a);

assume(t>0,ε>0,γ>0,R>0)

pa:p13b;

1e) limit  R→ ∞

pa1:limit(pa,R,∞);

2e) only when  ε → 0

pa2:limit(pa1,ε,0);

result proof::

p14:rhs(p2c) = pa2;

what about result for inverse laplace transform ?
=====================================
use :p2b

p15:p2b;

sum of the residues : poles enclosed by the contour ( C)
-----------------------------------------------------------------------------
C[i], m poles in the closed contour : i=1,2,...,m

FORMULA (1)
--------------------

p15a:lhs(p15) = 2*π*%i*sum(C[i],i,1,m);

p15b:rhs(p15a) =lratsubst([p14],rhs(p15));

p15c:p15b/rhs(p1a);

FORMULA (2)
--------------------

p15d:rhs(p15c) = lhs(p15c);