Program fails for lower values of g ; fsolve gives tolerance error ; integral dont converge

i want to calculate de qQd value, for a lower value of g, like this :

%angulo espalhamento

function exem6
  tic

    f=@(b) anggbsR(b);
    x0=[1,2];
    x=fzero(f,1.1);
    T=100;
    g=0.2;

    f1=@(b) 2*(1 + (sqrt(pi*T)/(2*g)).*sin(ang2(b)/2)).*b ;
    f2=@(b)  2*(1 - cos(ang2(b))).*b ;
    T=100;

    qQd = integral (f2,x, 10,'RelTol',0,'AbsTol',1e-7) + integral(f1,0,x) 

    toc

end

function Res = qQd(g)

    f=@(b) anggbsR(b);
    %x0=[1,2];
    x=fzero(f,1.1);
    %T=100;
    g=0.2;

    T=100; 
    f1=@(b) 2*(1 + (sqrt(pi*T)/(2*g)).*sin(ang2(b)/2)).*b ;
    f2=@(b)  2*(1 - cos(ang2(b))).*b ;
    Res = integral (f2,x, 10,'RelTol',0,'AbsTol',1e-7) + integral(f1,0,x);
  end

function xx2 = ang2(b)
  for i=1:length(b)
    xx2(i)=anggbsR(b(i));
  end
end

function xx=anggbsR(b)
  g=0.2;
  T=100;
  s= 1; 
  R= 2; 
  syms r
  func =  @(r)  1 - (b/r)^2 - (g^-2)*(2/15*[(s/R)]^9 *(1/(r - 1)^9 - 1/(r + 1)^9 - 9/(8* r)*(1/(r - 1)^8 - 1/(r + 1)^8)) - [(s/R)]^3 *(1/(r -1)^3 - 1/(r + 1)^3 - 3/(2* r)* (1/(r - 1)^2 - 1/(r + 1)^2)));
  minrootgbsR = fzero(func,1.1);
  k=1;
  for i=1:length(minrootgbsR)
    if(isreal(minrootgbsR(i))==1)
      vec(k)=minrootgbsR(i);
      k=k+1;
    end
  end

    fun =@(r) 1./(r.^2 .* sqrt(1 - (b./r).^2 - (g^-2)*(2/15*(s/R)^9 *(1./(r - 1).^9 - 1./(r + 1).^9 - 9./(8* r).*(1./(r - 1).^8 - 1./(r + 1).^8)) - (s/R)^3 *(1./(r -1).^3 - 1./(r + 1).^3 - 3./(2* r).* (1./(r - 1).^2 - 1./(r + 1).^2))))); 
    xx = real ( pi - 2*b * integral(fun,max(vec),Inf));

end

for g=0.02 , this integral doesnt work, but for g=10, for example, works great. fsolve gives a tolerance error .. fzero dont work. What Happened ? solver integral

[Walter Roberson]

Note: in anggbsR you have

syms r

but you never use r as a symbol. When you use

func = @(r) .....

then r because a local parameter just as if you had written

function value = func(r)
   ...
end

and it "shadows" the meaning of r as a symbol. Another term for this is "dummy variable"

[Walter Roberson]

I had to do a lot of clean up of the indentation to make it readable. Please use the "smart indentation" feature of the MATLAB code editor on your code. It was difficult to tell where some of your variables where coming from because of the indentation you had used.