Partial integration of the function f = 1

[caleb5040]

Given a function z(x,y), is the following expression valid:

∫(1)∂z = z

Does this even make sense? I came up with this in my differential equations class, but I'm not sure if it actually means anything.

 

[HallsofIvy]

Correct except that you do not have the "constant of integration". Which, because the partial derivative with respect to z treats other variables as constants, may be a function of those other variables. That is, if your variables are x, y, and z, the "anti-derivative" of 1, with respect to z, is z+ f(x,y) where f can be any function of two variables.

 

[caleb5040]

I think that makes sense. Thanks!

 

[HallsofIvy]

Note that if you have the three partial differential equations
\frac{\partial f}{\partial x}= 1
\frac{\partial f}{\partial y}= 1
\frac{\partial f}{\partial z}= 1

The last equation gives, as I said, f(x,y,z)= z+ g(x,y). Differentiating that with respect to y,
\frac{\partial f}{\partial y}= \frac{\partial g}{\partial y}= 1
so that g(x,y)= y+ h(x). Differentiating that with respect to x,
\frac{\partial g}{\partial x}= \frac{dh}{dx}= 1
which gives h(x)= x+ C. Putting all of those together, f(x,y,z)= z+ g(x,y)= z+ y+ h(x)a= z+ y+ x+ C.