Fwiw, it can be written in vector notation as ##\phi=\int \vec\nabla.\vec{dr}-\int\int (\vec \nabla\times\vec\nabla)\phi.(\vec{dr}\times\vec{dr})+\int\int\int \wedge^3\vec \nabla\phi.\wedge^3\vec{dr}##, where ##\wedge^3\vec v## stands for the triple scalar product ##\vec v.\vec v\times\vec v## (by analogy with the exterior product).TSny said:Yes, you are right. I was making the stupid mistake of thinking that taking the partial derivative of a function with respect to x, say, followed by integrating with respect to x would give back the function (up to a constant of integration). Certainly not!
Yes, that needs to be clarified. I take the integrals to be analogous to partial derivatives, so integration dx treats y and z as constants.TSny said:haruspex, I'm a little unsure of how to interpret the integrals in your expression. Suppose we have two independent variables ##x## and ##y## and we want to give meaning to the integral ##\int{(x+y)^2}dx##. Doing the integral with a ##u## substitution ##u = x + y## yields ##(x+y)^3/3##. But, instead, if I square out ##(x+y)^2## and then integrate, I get ##x^3/3 + x^2y + xy^2##. These results differ by ##y^3/3##, which can be considered a "constant of integration" since ##y## is held constant during the integration with respect to ##x##. So I'm having trouble seeing how the integrals in your expression lead to a definite result.
Yes, that's how I was interpreting the integrals. I just don't see how to decide on the "arbitrary functions of integration". I will think some more about it tomorrow.haruspex said:Yes, that needs to be clarified. I take the integrals to be analogous to partial derivatives, so integration dx treats y and z as constants.
What I was hoping to achieve is that you don't have to worry about the arbitrary functions. They should come in via the other integrals. All that should be arbitrary at the end is a constant. It seems to work for the cases I've tried.TSny said:Yes, that's how I was interpreting the integrals. I just don't see how to decide on the "arbitrary functions of integration". I will think some more about it tomorrow.