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complex contour integral ...plz help me

4 Ansichten (letzte 30 Tage)
adrian zizo
adrian zizo am 15 Mär. 2015
Kommentiert: Siwei Wang am 24 Apr. 2021
I write the limits of integration from 0 to 2*pi when he give me a Full Circle ,, but what I write
instead of 2*pi when he give me half circle as show in figure? :
this statement of matlab when he give me a contour for full circle :
q1=quad(@(t) fun(g(t)).*gprime(t),0,2*pi)
I want instead it by contour for half circle ,, what I write instead( 0,2*pi)?
this problem that I want answer it:
find the integral around half contour
∫(x^3+3)/((x^2+1)(x^2+4))

Antworten (1)

Torsten
Torsten am 16 Mär. 2015
fun=@(z)(x.^3+3)./((x.^2+1).*(x.^2+4));
g1=@(theta)R*cos(theta)+1i*R*sin(theta);
g1prime=@(theta)-R*sin(theta)+1i*R*cos(theta);
g2=@(t)-R+t*2*R;
g2prime=2*R;
q=quad(@(t)fun(g1(t)).*gprime(t),0,pi)+quad(@(t)fun(g2(t))*g2prime,0,1);
Best wishes
Torsten.
  1 Kommentar
Siwei Wang
Siwei Wang am 24 Apr. 2021
I correct a few typos in Torsten's answer:
R=10;
fun=@(x)(x.^3+3)./((x.^2+1).*(x.^2+4));
g1=@(theta)R*cos(theta)+1i*R*sin(theta);
g1prime=@(theta)-R*sin(theta)+1i*R*cos(theta);
g2=@(t)-R+t*2*R;
g2prime=2*R;
q=quad(@(t)fun(g1(t)).*g1prime(t),0,pi)+quad(@(t)fun(g2(t))*g2prime,0,1)
The result is q=1.57079596718350 + 3.14159265928277i

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