Integral Error message: first input argument must be a function handle

h= @(p)(p./DAK(0.6,p,150))
pseudop = integral(h,14.7,5000)

DAK is a function that's largely dependent on "p"(pressure), the other inputs are constant. The first line of code works fine but I keep getting an error for the second line of code. I tried fitting an equation with p as the variable for function DAK but the fit is bad and will negatively affect my solution..

 This is the error message: Error using integral (line 82) First input argument must be a function handle.

This doesn't work either because the program keeps running...

integral(@(p)(p./DAK(0.6,p,150)),14.7,5000)

[Birdman]

Can you also share your DAK function?

[Temitope Amao]

Here it is:

function Z = DAK(SGg,p,T) % gas compressibility factor

% Calculates the compressibility factor of natural gases for a range of pressures (Psia) 'minP' to
% 'maxP' in steps 'Pstep' (for a given temperature T(degree Farenheit) and specific gravity sg) using the Dranchuk-Abbou Kassem
% Equation of state. The range of validity is 0.2<Ppr<30 and 1.0<Tpr<3.0 (Craft
% &Hawkins, 1991)

%To calculate for a single pressure point P, set minP = maxP = Pstep
format short g
i = 1;
%for P = minP:Pstep:maxP % 
for P = p % 
    % Estimation of Psuedocritical pressure and temperature based on
    % suttons correlations (Craft & Hawkins, 1991)

      Ppc = 677+15*SGg-37.5*SGg^2;
      Tpc = 168+325*SGg-12.5*SGg^2;

      %Calculation of Pseudo reduced pressure and temperature
      Ppr = P / Ppc;
      Tpr = (T+459.67)/ Tpc;

      % Correlation Constants, c4 has to be updated within the root finding
      % algorithm
      A1 = 0.3265;
      A2 = -1.0700;
      A3 = -0.5339;
      A4 = 0.01569;
      A5 = -0.05165;
      A6 = 0.5475;
      A7 = -0.7361;
      A8 = 0.1844;
      A9 = 0.1056;
      A10 = 0.6134;
      A11 = 0.7210;
      c1 = A1 + (A2/Tpr) + (A3/(Tpr^3))+ (A4/(Tpr^4))+ (A5/(Tpr^5));
      c2 = A6 + (A7/Tpr) + (A8/(Tpr^2));
      c3 = A9*((A7/Tpr) + (A8/(Tpr^2)));

     % Implementation of bisection root finding method (The Newton-Raphson method becomes unstable and slower 
     % at low temperature and higher specific gravities, the bisection method
     % appears more robust for all cases in this specific application)
      brack1 = 1e-2;
      brack2 = 4;
      itr= 0;
      while 2<3 
          zguess = (brack1 + brack2) / 2;
          Rrguess = (0.27*Ppr) /(zguess*Tpr);
          c4 = (A10)*(1+(A11*(Rrguess^2)))*((Rrguess^2)/(Tpr^3))*(exp(-A11*(Rrguess^2)));
          syms z
          Rr = (0.27*Ppr) /(z*Tpr);
          f = z+ (c3*(Rr^5)) - (c2*(Rr^2)) - (c1*(Rr^1)) - c4 - 1;

          fzguess = subs(f,z,zguess);

          convergence  = abs(fzguess-0);
          conv(itr+1) = convergence;
          disp_convergence = conv'; % remove ";" to display convergence progression for each pressure point

          if convergence <=10^-4
              itr = itr + 1;
              break
          end

          if fzguess < 0
              brack1 = zguess;
          else
              brack2 = zguess;
          end

        itr = itr + 1;

      end
      Z(i) = zguess;
      Pressure(i) = P;
      I__________P__________Z_____________I = [Pressure' Z'];
      i=i+1;

%Uncomment following comments to view convergence progression for each
%pressure point

%     plot(disp_convergence)
%     xlabel('Convergence','fontsize',12)
%     ylabel('Iteration','fontsize',12)
%     hold on
%     pause

      clear conv
  end

% hold off

% figure;plot(Pressure,Z, 'r*','MarkerSize',6)
% xlabel('Pressure','fontsize',12);
% ylabel('Z','fontsize',12)

end