Calculating a three variable complex integral with Cauchy Teorem
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I'm trying to evaluate this complex three variable integral:
@(qx,qy,qz)(0.9383/((2*pi)^3*2*sqrt(qx^2+qy^2+qz^2+0.938^2)*sqrt(0.938^2+(qx^2+qy^2+qz^2)*(s-sqrt(qx^2+qy^2+qz^2+0.9383^2)-sqrt(0.9383^2+(qx^2+qy^2+qz^2)+000000.1*i)))))
This is the function I have created, which I have called "propagator_cauchy"
syms qx
syms qy
syms qz
syms s
C = [1+1i -1+1i -1-1i 1-1i]; %This is the integration curve
f=input('write the function you want to integrate: ') ;
q1=input('input first differencial: ');
i1=integral(f,q1,1,1,'Waypoints',C);
t=i1;
q2=input('input second differencial: ');
i2=integral(t,q2,1,1,'Waypoints',C);
p=i2;
q3=input('input third differencial: ');
i3=integral(p,q3,1,1,'Waypoints',C);
But when running this function, this is what appears:
>> propagator_cauchy
write the function you want to integrate: @(qx,qy,qz)(0.9383/((2*pi)^3*2*sqrt(qx^2+qy^2+qz^2+0.938^2)*sqrt(0.938^2+(qx^2+qy^2+qz^2)*(s-sqrt(qx^2+qy^2+qz^2+0.9383^2)-sqrt(0.9383^2+(qx^2+qy^2+qz^2)+000000.1*i)))))
input first differencial: qx
Error using integral (line 85)
A and B must be floating-point scalars.
Error in propagator_cauchy (line 12)
i1=integral(f,q1,1,1,'Waypoints',C);
May anyone know how can I solve this problem?
Thanks in advance
1 个评论
David Goodmanson
2020-1-11
编辑:David Goodmanson
2020-1-11
Hi Lara,
could you post the expressions that the code is based on, preferably with some information on the context? Rather than three separate contour integrals, it could be more likely that the integral is a single contour in the complex q^2 plane. Either way, terms like sqrt(q^2 + something) or sqrt(q1^2 + something) mean that the path of integration will run into branch cuts and mess up direct use of Cauchy's theorem. p.s. looks like the expression for i3 should say integral instead of int.
回答(2 个)
LARA VELASCO DAVOISE
2020-1-12
编辑:LARA VELASCO DAVOISE
2020-1-12
LARA VELASCO DAVOISE
2020-1-13
编辑:LARA VELASCO DAVOISE
2020-1-13
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