Integral change of variables formula confusion

[SchroedingersLion]

Greetings all.

I just got confused by the following.
Consider volume integral, for simplicity in 1D.
$$
V(A) = \int_{A} dz.
$$

If $z$ can be written as an invertible function of $x$, i.e. $z=f(x)$, we know the change of variables formula
$$
V(A)=\int_{A} dz= \int_{z^{-1}(A)} |z'(x)|dx.
$$

Intuitively, this is clear. What confused me now was the notion that the volume element transforms according to
$$
dz = |z'(x)|dx.
$$

If this equality holds, it should be allowed to write
$$
\int_{A} dz= \int_{A} |z'(x)|dx,
$$
which is obviously wrong. Why exactly is this not allowed?

 

[anuttarasammyak]

Are you wondering whether A or $z^{-1}(A)$ the domain of integration is ?
It would be A for $\int dz ...$ and $z^{-1}(A)$ for $\int dx ...$.

 

[fresh_42]

Greetings all.

I just got confused by the following.
Consider volume integral, for simplicity in 1D.
$$
V(A) = \int_{A} dz.
$$

If $z$ can be written as an invertible function of $x$, i.e. $z=f(x)$, we know the change of variables formula
$$
V(A)=\int_{A} dz= \int_{z^{-1}(A)} |z'(x)|dx.
$$

Intuitively, this is clear. What confused me now was the notion that the volume element transforms according to
$$
dz = |z'(x)|dx.
$$

If this equality holds, it should be allowed to write
$$
\int_{A} dz= \int_{A} |z'(x)|dx,
$$
which is obviously wrong. Why exactly is this not allowed?

Because written out, the integral reads $\displaystyle{\int_{z=a}^{z=b}}dz$, and you cannot substitute the variable in one place and ignore the other places. Here is the multivariate version:
https://en.wikipedia.org/wiki/Integration_by_substitution#Substitution_for_multiple_variables

 

[SchroedingersLion]

Because written out, the integral reads $\displaystyle{\int_{z=a}^{z=b}}dz$, and you cannot substitute the variable in one place and ignore the other places. Here is the multivariate version:
https://en.wikipedia.org/wiki/Integration_by_substitution#Substitution_for_multiple_variables

Thank you. Writing the integration domain explicitly in terms of the variable helps to clear my confusion. I need to replace everything $z$-related with the corresponding $x$ expression, not just a part of the whole thing.

 

[vanhees71]

Of course you have to change the integration range to the new variables. You also must make sure that the transformation is a diffeomorphism over the entire range of integration. For a 1D integral over an interval the function must be monotonous and differentiable verywhere along the interval. Then
$$\int_a^b \mathrm{d} z f(z) = \int_{z^{-1}(a)}^{z^{-1}(b)} \mathrm{d} u \frac{\mathrm{d}}{\mathrm{d}u} z(u)f[z(u)].$$