The Jacobian's negative, why plug in a positive number?

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[vernal sail]

(please ping when answering, ty!)

Update: I noticed the absolute value signs around the jacobian.

New question: why do we use the absolute value? Is it not possible that a change in (u,v) corresponds to a negative change in (x,y) :,)

[austere pond]

I have not done anything like this at all yet (in first year of undergrad) but would the absolute value be used to prevent the flipping of the region of the integral?? just a guess

@vernal sail

[plush adder]

Or, rather, to prevent considering signed areas

[vernal sail]

oh that's a good possibility hmm

[minor bane]

The theorem of the variable change indicates that you must use the absolute value of the determinant of the jacobian

while this sounds extremely long winded, it's quite related to how you measure sets with quantities that are always nonnegative

for instance, when you want to measure the area of a rectangle, it does not matter all that much if you count the width from left to right or from right to left

Though I'm not sure how familiar you are with measure theory, so there's probably a limit of how much I can explain without going in-depth

[vernal sail]

Yeah I don't understand much 😅

Is there a simple run down version? If not it's fine

[minor bane]

Let me try to think how to formulate this without overly confusing you

Would exemplifying work fine with you? Just showing that it is the case for very simple cases

@vernal sail

[vernal sail]

Yeah!

[minor bane]

Think of the following

$f$ is differentiable in $\mathbb{R}$ and you wanna look on a certain region (in $\mathbb{R}$ and Riemannian integration, this would correspond to intervals, so say in $(a, b)$)

[dusk matrix]

Rion

[plush adder]

Ah, f(x) = x^2 and u = -x would work

[minor bane]

$\int_{a}^{b} f'(x) dx = f(b) - f(a)$

[dusk matrix]

Rion

[minor bane]

And now, let's suppose that you do a variable change that is a) smooth, b) strictly decreasing

As Rafain suggested, $u = -x$ is perfectly sound

[dusk matrix]

Rion

[minor bane]

Method 1: Riemannian integration

$du = -dx$, and new bounds are $-a$ to $-b$

Hence:

[dusk matrix]

Rion

[minor bane]

$\int_{a}^{b} f'(x) dx = \int_{-a}^{-b} f'(-u) (-du) = \int_{-b}^{-a} f'(-u) du$

[dusk matrix]

Rion

[minor bane]

Notice here that the interval is written in the right order if a < b, since then -b < -a

Method 2: with Lebesgue integration

Here, your variable change is smooth and strictly monotonic $u = \phi(x) = -x$

[dusk matrix]

Rion

[minor bane]

So what you can do is to compute the new image, i.e. the new region where you will integrate

$\phi[(a, b)] = (-b, -a)$

[dusk matrix]

Rion

[minor bane]

Then, compute the absolute value of the determinant of the jacobian (here it's just 1D)

$\lvert \phi'(x) \rvert = \lvert -1 \rvert = 1$

[dusk matrix]

Rion

[minor bane]

So here, what you get in the end is that:

$\int_{a}^{b} f(x) dx = \int_{(a, b)} f(x) dx = \int_{(-b, -a)} f(-u) \lvert (\phi^{-1})'(u)\rvert du$

[dusk matrix]

Rion

[minor bane]

so in terms of determinant, the $\lvert (\phi^{-1})'(u)\rvert$ thing is nothing but: $\frac{1}{\lvert \phi'(\phi^{-1}(u))\rvert}$

[dusk matrix]

Rion

[minor bane]

Which, according to the statement above, is nothing but 1

Another more interesting example would be to try with the variable change u = -2x

Method 1: du = -2dx, dx = (-du/2)

$\int_{a}^{b} f(x) dx = \int_{-2a}^{-2b} f \left( -\frac{u}{2}\right) \frac{-du}{2} = \int_{-2b}^{-2a} f \left( -\frac{u}{2}\right) \frac{du}{2}$

[dusk matrix]

Rion

[minor bane]

Method 2: $\phi(x) = -2x$, hence $\lvert \phi'(x) \rvert = 2$. On the other hand, the image is $(-2b, -2a)$

[dusk matrix]

Rion

[minor bane]

So:

$\int_{(a, b)} f(x)dx = \int_{(-2b, -2a)}\left( -\frac{u}{2}\right) \frac{du}{ \lvert \phi'(\phi^{-1}(u)) \rvert} = \int_{(-2b, -2a)} f\left( -\frac{u}{2}\right) \frac{du}{2}$

[dusk matrix]

Rion

[minor bane]

same same. In a sense, the minus sign was already accounted for when computing the new domain of integration

You can try with more elaborate (smooth and bijective) variable changes like $\phi(x) = 4x^3$

[dusk matrix]

Rion

[minor bane]

Or, in two dimensions, the infamous: $\phi^{-1}(r, \theta) = (r \cos \theta, r \sin \theta)$

[dusk matrix]

Rion