Take integral of $\int 1/x \,dz$ where $z=x+y$?

Question

I want to integrate

$$\int \frac{1}{x} \,dz$$

where the measure is the total differential of $z=x+y$ and $x,y\in\mathbb{R}$ are variables. I wonder if the result should simply be:

$$\int \frac{1}{x}\, dz = \int \frac{1}{x} \,dx+\int \frac{1}{x} \,dy = C_1+\log(x)+\frac{y}{x}$$

This seems too easy to be true. How should I compute this integral properly?

Thanks for any suggestion!

Answer 1

When $x$ and $y$ are independent variables the $1$-form $$\omega:={1\over x}(dx+dy)$$ is not exact, since ${\rm curl}\left({1\over x},{1\over x}\right)=-{1\over x^2}\not\equiv0$. This means that there is no function $f:\>\Omega\to{\mathbb R}$ defined in some domain $\Omega\subset{\mathbb R}^2$ such that $\omega=df$, which then could serve as a kind of "indefinite integral" in your example.

What makes sense however are line integrals along curves $$\gamma:\quad [a,b]\to\Omega,\qquad t\mapsto\bigl(x(t),y(t)\bigr)\ .$$ For such a curve the integral $$\int_\gamma\omega=\int_\gamma {1\over x}(dx+dy)=\int_a^b{1\over x(t)}\bigl(x'(t)+y'(t)\bigr)\>dt$$ is well defined, but its value depends on the full data inherent in $\gamma$, and not only on the endpoints of $\gamma$.

Answer 2

We have that $$z = x + y \implies x = z - y$$ Then the integral becomes $$\int\frac1x\mathrm dz = \int\frac1{z - y}\mathrm dz = \ln(z - y) + C$$ since $y$ is constant with respect to $z$.