Help me prove integral answer over infinitesimal interval

In the book, I see the following:

##\int_{x_1}^{x_1 + \epsilon X_1} F(x, \hat y , \hat y') dx = \epsilon X_1 F(x, y, y')\Bigr|_{x_1} + O(\epsilon^2)##.

My goal is to show why they are equal. Note that ##\hat y(x) = y(x) + \epsilon \eta(x)## and ##\hat y'(x) = y'(x) + \epsilon \eta'(x)## and ##\epsilon## is infinitesimal.

##\int_{x_1}^{x_1 + \epsilon X_1} F(x, \hat y , \hat y') dx = \int_{x_1}^{x_1 + \epsilon X_1} F(x, y + \epsilon \eta , y' + \epsilon \eta') dx##

Now, I use taylor around ##\epsilon = 0##. So:

##\int_{x_1}^{x_1 + \epsilon X_1} F(x, \hat y , \hat y') dx = \int_{x_1}^{x_1 + \epsilon X_1} [F(x, y, y') + \epsilon \eta \frac{\partial F}{\partial y} + \epsilon \eta \frac{\partial F}{\partial y'}] dx + O(\epsilon^2)##

Note that the 2nd and 3rd members of the integrand contains ##\epsilon## and when integrated over ##x_1## to ##x_1 + \epsilon X_1##, this will bring another ##\epsilon## and they will become the order of ##\epsilon^2##. So, we have:

##\int_{x_1}^{x_1 + \epsilon X_1} F(x, y, y') dx+ O(\epsilon^2)##

Now, sure, since ##\epsilon## is infinitesimal, we can say that over such small interval, ##F## doesn't change much and on that interval, it's always the same and we can take its value at ##x_1##, so we end up with what we set out to prove.Question 1: is my analysis correct ?

Question 2: See the Book(page 147). Note that author doesn't use the approximation sign and uses the equality directly. Does he mean that they're strictly equal or should he have used the approximation ? I need you to be sure about this, because this exactly determines if I understand things correctly. If it's exact equality and it's supposed to be like that, then my analysis is incorrect.

It is better to expand [tex]
I(\epsilon) = \int_{x_1}^{x_1 + \epsilon X_1} F(x, y + \epsilon \eta, y' + \epsilon \eta')\,dx[/tex] in a Taylor series directly, so that [tex]
I(\epsilon) = I(0) + \epsilon I'(0) + O(\epsilon^2).[/tex] There is no need for an [itex]\approx[/itex] sign; the statement is already approximate due to the presence of the [itex]O(\epsilon^2)[/itex] term.

Now [itex]I(0) = 0[/itex] is obvious and you can use the general result [tex]
\frac{d}{dx} \int_a^{g(x)} f(x,t)\,dt = g'(x)f(x,g(x)) + \int_a^{g(x)} \frac{\partial f}{\partial x}\,dt[/tex] to compute [itex]I'(0)[/itex].

Thanks for the answer @pasmith .

1. can you tell what was wrong in my analysis ?

2. can you tell why ##I(0) = 0## ?

3. I know that it's already an approximation, but assume that ##O(\epsilon^2)## is fully represented by the actual value. By approximation, I meant that when author makes it equal to ##\epsilon X_1 F(x, y, y')\Bigr|_{x_1}##, that's an approximation and clearly needs that sign instead of equality sign.

gionole said:

Thanks for the answer @pasmith .

1. can you tell what was wrong in my analysis ?

Why do you think that something is wrong as far as it went?

gionole said:

2. can you tell why ##I(0) = 0## ?

It's the integral from ##x_1## to ##x_1##. Any zero length integral (except for the Dirac delta function) is 0.

gionole said:

3. I know that it's already an approximation, but assume that ##O(\epsilon^2)## is fully represented by the actual value. By approximation, I meant that when author makes it equal to ##\epsilon X_1 F(x, y, y')\Bigr|_{x_1}##, that's an approximation and clearly needs that sign instead of equality sign.

The ##O(\epsilon^2)## should be thought of as whatever will make an exact equality in each equation where it appears, not necessarily the same thing in different equations. The notation indicates that, whatever the exact term is, it is of order ##\epsilon^2##.

I think in my calculation, I was not making ##\epsilon## to be 0 in the upper limit of the integral which is what I meant by "what's the mistake in my calculation". I clearly see this was a mistake that I had.

If you have ##I(\epsilon) = \int_{x_1}^{x_1 + \epsilon X_1} F(x, y + \epsilon \eta , y' + \epsilon \eta') dx##, then for sure, ##I(0) = 0##.

But the way I calculate ##\epsilon I'(0)## is, that's gonna be ##\epsilon \int_{x_1}^{x_1} \eta \frac{\partial F}{\partial y} + \eta \frac{\partial F}{\partial y'}] dx## and this also ends up 0. Doesn't it ?

gionole said:

I think in my calculation, I was not making ##\epsilon## to be 0 in the upper limit of the integral which is what I meant by "what's the mistake in my calculation". I clearly see this was a mistake that I had.

If you have ##I(\epsilon) = \int_{x_1}^{x_1 + \epsilon X_1} F(x, y + \epsilon \eta , y' + \epsilon \eta') dx##, then for sure, ##I(0) = 0##.

But the way I calculate ##\epsilon I'(0)## is, that's gonna be ##\epsilon \int_{x_1}^{x_1} \eta \frac{\partial F}{\partial y} + \eta \frac{\partial F}{\partial y'}] dx## and this also ends up 0. Doesn't it ?

For the Taylor series that @pasmith gave you, the items ##I(0)## and ##I'(0)## are evaluated exactly for ##\epsilon = 0## and the others allow ##\epsilon## to be a variable. ##I(\epsilon) = I(0) + \epsilon I'(0) + O(\epsilon^2)##

I am not sure i understand it. It's mind boggling how ##I(0) = 0##, but ##I'(0) \neq 0##.

gionole said:

I am not sure i understand it. It's mind boggling how ##I(0) = 0##, but ##I'(0) \neq 0##.

Consider a simple example:$$I\left(x\right)\equiv\int_{0}^{x}dz=x\Rightarrow I'(x)=1\Rightarrow I\left(0\right)=0\text{ and }I'(0)=1\neq0$$Is that "mind boggling"?

gionole said:

I am not sure i understand it. It's mind boggling how ##I(0) = 0##, but ##I'(0) \neq 0##.

Suppose ##I(\epsilon) = \epsilon##. Then ##I(0) = 0## and ##I'(0) = 1##.
It seems like you might want to brush up on single variable derivatives a little before you try to tackle the problem you posted.

(@renormalize just beat me to this.) :-)

I think you didn't get me. What I said was is this:

we say ##I(0) = 0## because instead of ##\epsilon##, we insert 0 (even in the upper limit of the integral) and integral becomes from ##x_1## to ##x_1##.

as for ##I'(0)##, this is basically saying: derivate ##I## with respect to ##\epsilon## which when I do, I get:
##\int_{x_1}^{x_1+\epsilon X_0} \eta \frac{\partial F}{\partial y} + \eta \frac{\partial F}{\partial y'}] dx## and now, inserting ##\epsilon## to be 0 also makes the integral to be from ##x_1## to ##x_1## and such integral is also 0. This is what I meant, but now I realize where my mistake lies.

derivating ##I## with respect to ##\epsilon## is not ##\int_{x_1}^{x_1+\epsilon X_0} \eta \frac{\partial F}{\partial y} + \eta \frac{\partial F}{\partial y'}] dx## but it's what what @pasmith said in reply ##2##, which starts to make sense now.

Will think a little bit more today and let you know tomorrow. Thanks all <3