Curve fitting with integral involved

Hello. I am having trouble generating an equation to estimate parameters given a set of data for B and T. I am about to use lsqcurve fitting but for some reason i could not make the code run.

The equation should be in the form of :

See below:

T = [120,210,400,540,650];
B = [-344,-157,-17.5,-0.85,10.5];

fun = @(constant,T)(int(((2.*pi.*(constant(1)).^3)./3).*(1-exp((-4.*constant(2)./T)*((1./y.^4)-(1./y.^2)))),y,0,10));

T0 = [110,200]; %I just entered a random number, but don't know what this is for
T = lsqcurvefit(fun,T0,T,B);

The error I am getting is:

Unrecognized function or variable 'y'.

Error in

samplesept19>@(constant,T)(int(((2.*pi.*(constant(1)).^3)./3).*(1-exp((-4.*constant(2)./T)*((1./y.^4)-(1./y.^2)))),y,0,10))

(line 9)

fun =

@(constant,T)(int(((2.*pi.*(constant(1)).^3)./3).*(1-exp((-4.*constant(2)./T)*((1./y.^4)-(1./y.^2)))),y,0,10));

Error in lsqcurvefit (line 225)

initVals.F = feval(funfcn_x_xdata{3},xCurrent,XDATA,varargin{:});

Error in samplesept19 (line 12)

T = lsqcurvefit(fun,T0,T,B);

Caused by:

Failure in initial objective function evaluation. LSQCURVEFIT cannot continue.

Hoping for answers and help. :) Thank you so much!

Answer 1

[Walter Roberson]

format long g
T = [120,210,400,540,650];
B = [-344,-157,-17.5,-0.85,10.5];

syms t y
syms constant [1 2]
expr = ((2.*pi.*(constant(1)).^3)./3).*(1-exp((-4.*constant(2)./t)*((1./y.^4)-(1./y.^2))))

expr = 

intexpr = int(expr, y, 0, 10)

intexpr = 

RHS = 2/3.*pi.*(constant(1)).^3 .* intexpr;

fun = matlabFunction(RHS, 'vars', {constant, t})

fun = function_handle with value:
    @(in1,t)in1(:,1).^3.*pi.*integral(@(y)in1(:,1).^3.*pi.*(exp((in1(:,2).*(1.0./y.^2-1.0./y.^4).*4.0)./t)-1.0).*(-2.0./3.0),0.0,1.0e+1).*(2.0./3.0)

obj = @(P,T) arrayfun(@(t) fun(P,t), T)

obj = function_handle with value:
    @(P,T)arrayfun(@(t)fun(P,t),T)

P0 = [110,200]; %I just entered a random number, but don't know what this is for
P = lsqcurvefit(obj, P0, T(:), B(:))

Local minimum possible.

lsqcurvefit stopped because the final change in the sum of squares relative to 
its initial value is less than the value of the function tolerance.

P = 1×2

1.6038792022093 184.707909161448

B2_predicted = obj(P,T(:));
plot(T, B, 'k*', T, B2_predicted, 'b-+')
legend({'original B', 'predicted B'}, 'location', 'best')

[lvenG]

Hello Sir @Walter Roberson. Thank you so much for the help. Is there any way that this code can calculate the value of constant(1) and constant(2)? When I run this, it just returns the value constant and did not calculate it. Please enlighten me. Thank you.

[Walter Roberson]

P(1) is constant(1) and P(2) is constant(2)

[lvenG]

@Walter Roberson, thank you so much Sir. This worked.... Really, thanks for the help and enlightenment... Hope you are having a great day! Stay safe.