Error using integral: A and B must be floating-point scalars

Hey there,

I try to solve some equations where I have to use integrals. I'm not very experienced with function handles, which is where I assume to get the error.

function opt_barrier = derive_p3(beta, lambda, d)

popt = 0;
f=cell(100);
Copt=cell(100);

f0 = @(x,p) x-p+((lambda*beta*d-1)/((1-beta)*lambda));
f{1} = @(x,p) (beta*(beta + lambda*(d+(beta-1)*p)-2))/((1-beta)*lambda)+beta*x - ((1-beta)/lambda) *exp(lambda*p-lambda*x);
syms p;
syms Ctilde;

C= (beta-1)*exp(lambda*p)/lambda;
Copt{1} = C;
Vp= (lambda*beta*d-1)/((1-beta)*lambda);

p_cand=solve(exp(-lambda*(p+d))==((beta-1)^2+beta*(beta-1)^2*(-exp(-lambda*d))*(lambda*p-2))/(beta^2*(2*beta+lambda*(d+(beta-1)*p)-3)), p);
    if 0 <=p_cand & p_cand <= d
        popt = p_cand;
    else
    i=1;
        while popt == 0;

          integrand = @ (x,p) f{i}(x,p).*exp(lambda*x);
          Copt{i+1} = @(x,p) solve(integral(@(t) integrand(t+d,p),0, p-i.*d).*exp(-lambda.*(p-i.*d))+ Ctilde == f{i}(p-i.*d, p), Ctilde);
          helpfunc1 = @(x,p) exp(-lambda.*x);
          intop = @(x,p) integral(@(t) integrand(t+d,p),0,x);
          f{i+1} = @(x,p) intop(x,p)*helpfunc1(x,p)+ Copt{i+1}(x,p);

          term= @(x,p)0;

          for j=1:i
             term = @(x,p) term(x,p) + integral(@(y) f{i}(y,p).*lambda.*exp(lambda.*(y-d-x)),p-j.*d, p-(j-1).*d);
          end

          x=p-i.*d;

          helpfunc2 = @(y,p) lambda.*exp(lambda.*(y-d-x));
          integrand1 = @(y,p) f{i+1}(y,p)*helpfunc2(y,p);
          integrandlast = @(y,p) f0(y,p)*helpfunc2(y,p);

          cand_p = solve(f{i+1}(x,p)== beta.*integral(@(y)integrand1(y,p), 0, 1) +  beta.*term(x,p) + beta.*int(integrandlast,y, p, d+x), p)

          popt=1;

          end

      end

      opt_barrier=popt;
  end

In the while-loop, I get the error 'Error using integral:A and B must be floating-point scalars' where I calculate p_cand, the rest is working. I assume that the integral function has problems with the 'syms' of p, but I don't know how to solve equations without using syms.

Thanks for your help!

[Arif Ullah khan]

term = @(x,p) term(x,p) + integral(@(y) f{i}(y,p).*lambda.*exp(lambda.*(y-d-x)),p-j.*d, p-(j-1).*d);

p must be a scalar value in order to make the limits of integral scalar. but you defined it as symbolic variable

syms p;

if you are doing symbolic integration then int function will work instead of integral

Answer 1

[Karan Gill]

To represent integration with symbolic variables, use "int" instead of "integral".

https://www.mathworks.com/help/symbolic/int.html