Re: Differentiation of Line Integral

"Kaba" <none@xxxxxxxx> wrote in message
news:MPG.1f4683fd53636e7b989782@xxxxxxxxxxxxxxxxx
That is the same formula as given by Hans Lamecker. It is missing the
absolute value over the Jacobian 1/(a-b).

No, it is not missing the Jacobian. See below.

I said it's missing the abs() function from around the Jacobian. That is
the definition I have used to. See below.

I see that I accidentally interchanged the limits of integration
when I posted. After the change of variables from Lamecker's
k to h, the integral should be
F(b) = int_{a}^{b} f(h) ((h-b)/(a-b)) dh/(a-b)
Still no absolute value.

Consider a definite integral of a 1-dimensional function f of a
1-dimensional variable x,
I = int_{a}^{b} f(x) dx
where a < b. You make the change of variables x = g(t),
where g(t) is differentiable and strictly monotonic, then
dx = g'(t) dt [no absolute values here]. Let a = g(c) and
b = g(d).
I = int_{c}^{d} f(g(t)) g'(t) dt

I mentioned that g is strictly monotonic. If g'(t) > 0 for
all t, then c > d. If g'(t) < 0 for all t, then c < d. If you
must include absolute value signs, then
I = int_{min{c,d}}^{max{c,d}} f(g(t)) |g'(t)| dt

Here is the generalization I meant to simplify in there that
unifies the task of changing variables in an integral,
whether a line integral or some other:
<snip>

Being a mathematician, I am well aware of these concepts.
My usage of the term "change of variables" regarding
integration, though, has been restricted to functions
g: R^n -> R^n for which the Jacobian matrix is not singular.

In your discussion with g : R^m -> R^n when m < n, I view
the problem as having an m-dimensional manifold embedded
in n-dimensional space. The integration is formulated
abstractly in the n-dimensional space, but to compute it,
you need to parameterize the manifold and obtain an
integration problem in m-dimensional space.

dR = |sqrt(det(g'(p)^T g'(p)))| dP

Of course, you do not need the absolute value signs here :)
And the usual assumptions on the metric tensor
g'(p)^T g'(p) is that it is positive definite, so the determinant
is positive, in which case you do not need to worry about
taking the square root of a negative number.

Although interesting, the change-of-variables discussion is off
the path that the OP is interested.

--
Dave Eberly
http://www.geometrictools.com


.

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