symbolic integration depends on different equivalent forms of function

I performed the following equaivalent symbolic integrations:

syms x y
A = int(x+y,x);
A_ = expand(A);         % just expanded (equaivalent) form of A
B  = expand(int(A,y))
B_ = expand(int(A_,y))

with the following results:

A =
(x*(x + 2*y))/2

A_ =
x^2/2 + y*x

B =
x^3/8 + (x^2*y)/2 + (x*y^2)/2

B_ =
(x^2*y)/2 + (x*y^2)/2

I expect B equal to B_, but there is a misterious additional term x^3/8 at B ??!!

Is that a bug???

Answer 1

[Torsten]

The difference is a usual "constant of integration".

If you differentiate both B and B_ with respect to y and then with respect to x, you'll arrive at the expression x+y in both cases:

syms x y
B_ = (x^2*y)/2 + (x*y^2)/2;
B = x^3/8 + (x^2*y)/2 + (x*y^2)/2;
A_ = diff(B_,y);
A = diff(B,y);
expr1 = diff(A_,x)

expr1 = 

expr2 = diff(A,x)

expr2 = 

[Michal]

Why this effect of integration constant depends on presence of expand() function at

A_ = expand(A)

???

[Torsten]

Because deep in the MATLAB rules for symbolic integration, the sum of two polynomial expressions (x^2/2 + y*x) is integrated differently from a single product polynomial ( (x*(x + 2*y))/2 ).

In short: Only the developers know the answer.

[Michal]

Yes, and this is the problem... Two different results of integration for two mathematically equivalent SIMPLE expressions!

The fact that backward double differentiation produces the same result is irrelevant.

[Torsten]

The fact that backward double differentiation produces the same result is irrelevant.

No. You try to determine a function for which

d/dx (d/dy(f(x,y)) = x+y. 

There are infinitely many functions for which this is true.

E.g. if you want to determine a function with d/dx (d/dy (f(x,y)) = 0, every function of the form

f(x,y) = f1(x) + f2(y) 

is "correct" .

If you want to make the result of integration unique, you must fix f(x,y) on a 1d-curve in the x-y-plane (which is not parallel to one of the coordinate axes, I guess).

[Michal]

The main point of my question is the fact, that

int(int(x+y,x),y)

is not same as

int(expand(int(x+y,x),y))

which is really strange!!! That is all ... But, you are right that: Only the developers know the answer.:)

[Torsten]

I'm not surprised that

int((x*(x + 2*y))/2,y)

gives a result different from

int(x^2/2 + y*x,y).

See

syms x
int((x-1)^2,x)

ans = 

compared to

int(x^2-2*x+1,x)

ans = 

[Paul]

I'm not surprised either. I've seen cases where int() couldn't find a solution unless the integrand was manipulated using simplify, expand, etc. Here is an example from this Question

syms t
assume(t, "real")
f1 = 3*t-t*t*t;
f2 = 3*t*t;
f = [f1, f2];
df = diff(f, t);
a = 0;
b = 1;
normDf = sqrt(df(1)*df(1)+df(2)*df(2));
int(normDf,t,a,b)  % no solution

ans = 

normDf = simplify(normDf)

normDf = 

int(normDf, t, a, b)  % easy

ans = 

4

Answer 2

[Michal]

Simple solution to avoid integration constant effect:

    int(f(x),x,0,x)

[Torsten]

This assumes that you want a special antiderivative F of f, namely the antiderivative with F(0) = 0. As I wrote, if in 1d, you fix the antiderivative in one point and in 2d, you fix the antiderivative on a curve, you get a unique solution for d/dx (d/dy (f(x,y))) = x+y.

[Michal]

int(int(f(x,y),y,0,y),x,0,x)

reliable produce results without misterious integration constant terms and does not depends on any pre-processing steps (like expand, simplify, etc.)

[Torsten]

syms x y
A1 = int(sin(x)+sin(y),x,0,x);
B1 = int(A1,y,0,y)

B1 = 

simplify(B1)

ans = 

A2 = int(sin(x)+sin(y),x);
B2 = int(A2,y)

B2 = 

You see, the integration constant of your method is x+y.

All I want to say is:

Each integration gives its individual integration constant. For the differential equation

d/dx (d/dy(F(x,y))) = x+y

e.g., the "integration constants" are functions of the form

G(x,y) = g1(x) + g2(y)

unless you fix F on a curve in the (x,y) plane (not parallel to one of he coordiante axes).

Some integration constants look more plausible, others less.

And now I will be quiet and you can have "the last word", if you want.

[Michal]

Well... You are right!