Exponent error calculating integral.

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Shubhankar Jape
Shubhankar Jape 2021-5-24
评论： Shubhankar Jape 2021-5-24
I am trying to calculate the intergral of the function given below.
clc
clear all
syms rho
format long
format compact
lambda = 100;
z_0 = lambda/100; t = lambda/100;
z_1 = z_0 + t;
limit = 1000:1000:5000;
%rhol = 1:100000;
y0 = lambda/3;
for i = 1:5
fun = @(phi) cos(2*pi.*y0./lambda).*(1./(phi^3)).*besselj(0,2.*pi./lambda.*(1./phi)).*log((-z_0+sqrt(z_0.^2+(1./phi).^2))./(-z_1+sqrt(z_1.^2+(1./phi).^2)));
fun_int(i) = integral(fun,0,limit(i));
end
%t=[1:1000];
plot(limit,fun_int)
%rho = logspace(0,5,10000);
%fun_rho = fun(rho);
%plot(rho,fun_rho)
%plot(t,fun_int);
%set(gca,'xscale','log');%set(gca,'yscale','log');
%ylabel('Function(\rho)');
xlabel('phi limit');
ylabel('f(phi)');
It throws an error
Error using ^ (line 51)
Incorrect dimensions for raising a matrix to a power. Check that the matrix is square and the power is
a scalar. To perform elementwise matrix powers, use '.^'.
Error in
diode_semilog_01>@(phi)cos(2*pi.*y0./lambda).*(1./(phi^3)).*besselj(0,2.*pi./lambda.*(1./phi)).*log((-z_0+sqrt(z_0.^2+(1./phi).^2))./(-z_1+sqrt(z_1.^2+(1./phi).^2)))
(line 15)
fun = @(phi)
cos(2*pi.*y0./lambda).*(1./(phi^3)).*besselj(0,2.*pi./lambda.*(1./phi)).*log((-z_0+sqrt(z_0.^2+(1./phi).^2))./(-z_1+sqrt(z_1.^2+(1./phi).^2)));
Error in integralCalc/iterateScalarValued (line 314)
fx = FUN(t);
Error in integralCalc/vadapt (line 132)
[q,errbnd] = iterateScalarValued(u,tinterval,pathlen);
Error in integralCalc (line 75)
[q,errbnd] = vadapt(@AtoBInvTransform,interval);
Error in integral (line 88)
Q = integralCalc(fun,a,b,opstruct);
Error in diode_semilog_01 (line 16)
fun_int(i) = integral(fun,0,limit(i));
Does anybody know how to fix this?

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回答（1 个）

the cyclist
the cyclist 2021-5-24
I expect you want
phi.^3
instead of
phi^3
inside your function fun

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