Symbolic matlabFunction() generates a non-working function for my problem

I want to convert the symbolic expression to a MATLAB function, but the result of matlabFunction() leads to a non-working function with an unspecified parameter t:

syms t real
l = [0;0;t]
syms x y z real
r = [x;y;z];
sep = r-l
dl = diff(l,t)
integrand = simplify(cross(dl,sep)./norm(sep).^3,"Steps",100)
Bsym = int(integrand,t,[0 Inf])
fcn = matlabFunction(Bsym,'File','fcn1');

I received:

function Bsym = fcn1(x,y,z)

%FCN1

% Bsym = FCN1(X,Y,Z)

% This function was generated by the Symbolic Math Toolbox version 25.2.

% 10-Mar-2026 12:09:20

t2 = abs(y);

t3 = x.^2;

t4 = y.^2;

t5 = z.^2;

t7 = -z;

t6 = t2.^2;

t8 = -t3;

t9 = -t4;

t10 = t+t7;

t11 = -t6;

t12 = t10.^2;

t13 = t3+t6;

t16 = t8+t9;

t14 = t5+t13;

t17 = t8+t11;

t18 = t12+t13;

t20 = sqrt(t16);

t19 = 1.0./sqrt(t14);

t23 = t20+z;

t27 = 1.0./sqrt(t18);

t21 = t19.*z;

t30 = t10.*t27.*x;

t31 = t10.*t27.*y;

if ~all(cellfun(@isscalar,{t23,t8,y,z}))

error(message('symbolic:sym:matlabFunction:ConditionsMustBeScalar'));

end

if (~(0.0 <= t23))

t0 = -(y.*(t21+1.0))./t13;

elseif ((0.0 <= t23) & (y ~= 0.0))

t0 = -y./t13-limit(t31,t == z+sqrt(t17),Left)./t13+limit(t31,t == z+sqrt(t17),Right)./t13+(t7.*t19.*y)./t13;

elseif ((0.0 <= z+sqrt(t8)) & (y == 0.0))

t0 = 0.0;

else

t0 = NaN;

end

if ~all(cellfun(@isscalar,{t23,t9,x,z}))

error(message('symbolic:sym:matlabFunction:ConditionsMustBeScalar'));

end

if (~(0.0 <= t23))

t1 = (x.*(t21+1.0))./t13;

elseif ((0.0 <= t23) & (x ~= 0.0))

t1 = x./t13+limit(t30,t == z+sqrt(t17),Left)./t13-limit(t30,t == z+sqrt(t17),Right)./t13+(t21.*x)./t13;

elseif ((0.0 <= z+sqrt(t9)) & (x == 0.0))

t1 = 0.0;

else

t1 = NaN;

end

Bsym = [t0;t1;0.0];

end

But if you try, you'll get an error.:

fcn1(1,1,1)

Unrecognized function or variable 't'.

Error in fcn1 (line 16)

t10 = t+t7;

^^^^^^^^^^^

Why did matlabFunction() skip the t parameter?

Answer 1

[Walter Roberson]

syms t real
l = [0;0;t]

l = 

syms x y z real
r = [x;y;z];
sep = r-l

sep = 

dl = diff(l,t)

dl = 

integrand = simplify(cross(dl,sep)./norm(sep).^3,"Steps",100)

integrand = 

Bsym = int(integrand,t,[0 Inf])

Bsym = 

symvar(Bsym)

ans = 

In Bsym, t appears only as a variable in a limit() call.

matlabFunction() cannot deal with limit properly, as limit() is inherently a symbolic call that cannot be translated to pure numeric functions.

[Walter Roberson]

Your equations have the possibility that x and y might be 0, in which case the denominator becomes ((t-z)^2)^(3/2) . But if z is positive then because t runs from 0 to inf, at some point t-z will be 0, leading to an all-zero denominator, which you will have trouble integrating over.

Note by the way, that with x and y both real, that involves the square root of a quantity that is at most 0 (if x and y are both 0) and is otherwise negative. So involves mostly complex quantities. The limit() calls generated involve conditionals on which is true if and only if x and y are both 0, and otherwise is a comparison to a complex quantity which is fallse. matlabFunction() really doesn't do that well on limit() calls.

[Paul]

"matlabFunction() cannot deal with limit properly,"

Why can't matlabFunction just punt with a message that "limit" can't be in the input expression?

[Torsten]

@Erik

Maybe you can directly use the antiderivatives:

syms x y z t
int(-y/((t-z)^2+y^2+x^2)^(3/2),t)

ans = 

int(x/((t-z)^2+y^2+x^2)^(3/2),t)

ans =