Integration with different infinitesimal intervals

[JohnnyGui]

TL;DR Summary

Difficulty grasping integration over infinitesimal intervals and their relation with the infinitesimal used for the integration itself.

Some sources state a similar format of the following
$$\int_a^{a+da}f(x)dx=f(a)da$$
Which had me thinking whether the following integration can exist
$$\int_a^{a+dx}f(x)dx=f(a)dx$$
I have difficulty grasping some aspects about these integrations

1. Regarding the 1st integration, shouldn't $a$ be a definite value of $x$ and not a variable? If yes, what is then the meaning of $da$? E.g. if $a=5$ then in my head I'd see $\int_5^{5+d5}f(x)dx=f(5)d5$ which doesn't make sense.

2. If the 2nd integration makes any sense, is it correct to say $da≠dx$?

3. If $da≠dx$, would there be a function $a(x)$ such that $f(a)dx=f(a)x'(a)da$? I have difficulty accepting this since $a$ is actually a value of $x$ according to the integration while this shows that $a$ is a function of $x$.

Can anyone help me clarify this?

 

[Mark44]

TL;DR Summary: Difficulty grasping integration over infinitesimal intervals and their relation with the infinitesimal used for the integration itself.

Some sources state a similar format of the following
$$\int_a^{a+da}f(x)dx=f(a)da$$
Which had me thinking whether the following integration can exist
$$\int_a^{a+dx}f(x)dx=f(a)dx$$
I have difficulty grasping some aspects about these integrations

1. Regarding the 1st integration, shouldn't $a$ be a definite value of $x$ and not a variable? If yes, what is then the meaning of $da$? E.g. if $a=5$ then in my head I'd see $\int_5^{5+d5}f(x)dx=f(5)d5$ which doesn't make sense.

Right. The first integral shouldn't have da as part of its upper limit of integration. Presumably a is a specific, but not necessarily known, value of x, so the upper limit should be a + dx.

2. If the 2nd integration makes any sense, is it correct to say $da≠dx$?

Not in my opinion.

3. If $da≠dx$, would there be a function $a(x)$ such that $f(a)dx=f(a)x'(a)da$? I have difficulty accepting this since $a$ is actually a value of $x$ according to the integration while this shows that $a$ is a function of $x$.

This doesn't make sense to me. Going back to the integral, is this what it would look like:
$\int_{a(x)}^{a(x) + da} f(x) dx$? Off the top of my head it doesn't look to me like this would simplify to f(a)da like you show in the first integral.

 

[WWGD]

Edit: In the Riemann Integral, you must use actual numbers in the outermost integral *or , if variables for an abstract argument, it must be variables representing numbers.

* Meaning if using iterated integrals.

 

[PeroK]

Can anyone help me clarify this?

Are you studying regular calculus or non-standard analysis?

 

[Frabjous]

‘x‘ is a dummy variable, so it should not be an integration limit.
The first integral is about the “area” interpretation of an integral.

 

[JohnnyGui]

Edit: In the Riemann Integral, you must use actual numbers in the outermost integral *or , if variables for an abstract argument, it must be variables representing numbers.

* Meaning if using iterated integrals.

Can I deduce from this that even if #a# is a variable for an abstract argument, then there's still no such thing as #da# because it must represent numbers? Or did I miss something here?

Right. The first integral shouldn't have da as part of its upper limit of integration. Presumably a is a specific, but not necessarily known, value of x, so the upper limit should be a + dx.

I don't get why it is stated differently here and here though. Can the limits be seen as a different variable while, at the same time, they are values of the integrand's variable? In that case I'd conclude that dx=da#.

Are you studying regular calculus or non-standard analysis?

Regular calculus, although I'm now curious if there's a non-standard analysis definition for this that can be explained to a layman?

‘x‘ is a dummy variable, so it should not be an integration limit.
The first integral is about the “area” interpretation of an integral.

So the 1st integration does make sense and the 2nd doesn't?

 

[Frabjous]

So the 1st integration does make sense and the 2nd doesn't?

I would say that the 1st makes sense, but is an abuse of notation. The second does not make sense.

 

[PeroK]

Regular calculus,

I would take a slightly different approach. $da$, in regular calculus, is a differential. The way it's defined is:

If $y = f(a)$, then $dy = f'(a)da$. We can also write $df = f'(a)da$. Now, for your integral, we can define a function:
$$F(a) =\int_0^{a}f(x)dx$$And note that$$F'(a) = f(a)$$Hence$$dF = F'(a)da = f(a)da$$Your source then takes $dF = \int_a^{a+da}f(x)dx$.

An even simpler way to think of it is that, for some small change $\Delta a$:
$$F(a + \Delta a) - F(a) = \int_a^{a+\Delta a}f(x)dx \approx f(a)\Delta a$$Then taking this to the differential limit, as it were, we have:$$F(a + da) - F(a) = \int_a^{a+da}f(x)dx \approx f(a)da$$

 

[fresh_42]

This entire discussion reminds me of a book I read in order to write
https://www.physicsforums.com/insights/when-lie-groups-became-physics/
This source was from 1911 when Lie theory (continuous transformation groups at the time) was developed. It is a bit old-fashioned and I would definitely stick to the modern treatment along either Riemannian (Darboux) integration or Lebesgue integration (measure theory) or answer ...

Are you studying regular calculus or non-standard analysis?

... first. I don't think that infinitesimals outside of non-standard analysis are very helpful anymore.

If they are used as in physics, then they represent basis vectors of a tangent or cotangent space.

 

[JohnnyGui]

I would take a slightly different approach. $da$, in regular calculus, is a differential. The way it's defined is:

If $y = f(a)$, then $dy = f'(a)da$. We can also write $df = f'(a)da$. Now, for your integral, we can define a function:
$$F(a) =\int_0^{a}f(x)dx$$And note that$$F'(a) = f(a)$$Hence$$dF = F'(a)da = f(a)da$$Your source then takes $dF = \int_a^{a+da}f(x)dx$.

An even simpler way to think of it is that, for some small change $\Delta a$:
$$F(a + \Delta a) - F(a) = \int_a^{a+\Delta a}f(x)dx \approx f(a)\Delta a$$Then taking this to the differential limit, as it were, we have:$$F(a + da) - F(a) = \int_a^{a+da}f(x)dx \approx f(a)da$$

Thank you. I'm more curious about the difference between $dx$ and $da$. Your explanation made me think of the following.
Can I say that in the case of
$$\int_a^{a+da}f(x)dx \approx f(a)da$$
...there is no relationship between $a$ and $x$ whatsoever ($x$ stays merely as an integrand variable) and also that $da\neq dx$?

 

[PeroK]

$x$ and $a$ are different variables. And $dx$ and $da$ are the differentials associated with those variables.

Can I say that in the case of
$$\int_a^{a+da}f(x)dx \approx f(a)da$$

If you mean $da$ as a small, finite change, then yes you havean approximation. This is usually written as $\Delta a$. If $da$ is a differential then you have equality - not an approximation.

 

[JohnnyGui]

x and a are different variables

So different variables that don't have a relation between them, since $x$ ultimately becomes $a$ after the integration?

Also, what is your stance on the upper limit being $a+dx$?

 

[PeroK]

So different variables that don't have a relation between them, since $x$ ultimately becomes $a$ after the integration?

$x$ is a dummy variable, so $a$ cannot meaningfully be a function of $x$.

Also, what is your stance on the upper limit being $a+dx$?

It's meaningless to use a dummy integration variable outside the integrand.

 

[JohnnyGui]

It's meaningless to use a dummy integration variable outside the integrand.

Thanks. I have one other question that came up when approaching this another way.
$f(x)dx$ is also equal to $dF(x)$. Now, if $x$ takes on a value $a$, then $dF(a)/dx=f(a)$. But we just saw that $dF(a)/da=f(a)$ as well.

Because $da$ is not equal to $dx$, such a similarity is only possible if $F$ is a straight or proportional line in the interval da-dx.

Is this statement correct and if not, where did I go wrong?

 

[pbuk]

This notation is hideous, no wonder it is confusing you.

From the definition of the Reimann integral, $\lim_{\delta x \to 0} \int_a^{a+\delta x} f(x) dx = \lim_{\delta x \to 0} f(a) \delta x$.

Is that any clearer?

 

[JohnnyGui]

This notation is hideous, no wonder it is confusing you. From the definition of the Reimann integral, $\lim_{\delta x \to 0} \int_a^{a+\delta x} f(x) dx = \lim_{\delta x \to 0} f(a) \delta x$. Is that any clearer?

Thank you, I'm trying but I fail to see how this answers whether my conclusion in post #14 is correct or not about getting the same $f(a)$ regardless of using $dx$ or $da$.

 

[Frabjous]

Thank you, I'm trying but I fail to see how this answers whether my conclusion in post #14 is correct or not about getting the same $f(a)$ regardless of using $dx$ or $da$.

There are variables and there are values. You are letting “a” be a variable and a value. Instead of writing down “a” write down a number, and then see what makes sense.

 

[PeroK]

Thanks. I have one other question that came up when approaching this another way.
$f(x)dx$ is also equal to $dF(x)$. Now, if $x$ takes on a value $a$, then $dF(a)/dx=f(a)$. But we just saw that $dF(a)/da=f(a)$ as well.

Because $da$ is not equal to $dx$, such a similarity is only possible if $F$ is a straight or proportional line in the interval da-dx.

Is this statement correct and if not, where did I go wrong?

You have some fundamental misconceptions about variables. In the original equation, $x$ is a dummy variable and $a$ is an independent variable.

Then you are taking $x$ and $a$ to be two independent variables.

Then you are using $a$ as a specific value of the variable $x$.

You need to sort out these different cases.

 

[WWGD]

@JohnnyGui , have you looked up how integration is developed within Non-Standard Analysis? If your " da" is an infinitesimal, that's the only way of dealing with it I believe makes sense here. Unfortunately, I'm not familiar with it myself.

 

[JohnnyGui]

@PeroK @Frabjous

Thank you, this cleared it up for me.

I have one last question because this source seems to say otherwise, or so I think.

It says that if
$$G(x)=\int_a^xf(t)dt$$
then one can deduce
$$G'(x)=f(x)$$
Which I can grasp. However, the poster also says that from... (note the $dt$ in the upper limit)
$$G(x+dt)-G(x)=\int_x^{x+dt}f(t)dt$$
One can also deduce $G'(x)=f(x)$

So the same $f(x)$ can be derived regardless of basing it on an interval $dt$ or $dx$? Or is the $x$ in the latter case also a value and not a variable?

 

[fresh_42]

@PeroK @Frabjous

Thank you, this cleared it up for me.

I'm not so certain considering the rest of your post.

I have one last question because this source seems to say otherwise, or so I think.

This source does not say anything like what you state in the rest of your post. It uses $h$ and $\Delta x$ in upper bounds, which is alright since these are real distances, nothing infinitesimal or differential.

It says that if
$$G(x)=\int_a^xf(t)dt$$
then one can deduce
$$G'(x)=f(x)$$
Which I can grasp.

If $F(x)$ is the anti-derivative of $f(x),$ i.e. $F'(x)=f(x),$ then
$$
\int_a^b f(t)\,dt=F(b)-F(a)
$$
by the fundamental theorem of calculus. This leads to
$$
\int_a^x f(t)\,dt=F(x)-F(a)
$$
if we consider the upper bound as a variable. Furthermore, if $G'(x)=f(x)$
$$
\left(\int_a^x f(t)\,dt \right)'=\dfrac{d}{dx}\left(\int_a^x f(t)\,dt \right)=(G(x)-G(a))'=G'(x)-G'(a)=G'(x)-0=G'(x)=f(x)
$$
This is true for any such function $G(x)$ since the differentiation makes constant additive terms as $G(a)$ or $F(a)$ zero. All that is needed is $G'(x)=f(x).$

However, the poster also says that from... (note the $dt$ in the upper limit)
$$G(x+dt)-G(x)=\int_x^{x+dt}f(t)dt$$
One can also deduce $G'(x)=f(x)$

See above. The poster said (with a lower bound $a=x$)
$$
\int_x^{x+\Delta X}f(t) \,dt= G(x+\Delta x) - G(x)
$$
and set $h=\Delta x$ but this is neither $dx$ nor $\delta x.$ It is a real number, a distance on the $x$-axis. We can now consider what happens if that distance $h$ tends to zero, but that doesn't allow us to write $dx$ in the upper bound. That's more than confusing. And $dt$ is doubly wrong.

So the same $f(x)$ can be derived regardless of basing it on an interval $dt$ or $dx$? Or is the $x$ in the latter case also a value and not a variable?

Read carefully!

The so-called dummy variable of integration is always $t$ as in $\int f(t)\,dt$ and never $dx.$
The bounds - upper or lower - are noted as the variable $x$ which is independent of integration variable $t$. Values of $x$ note points on the $x$-axis and $h=\Delta x$ note distances on the $x$-axis. It is investigated what happens if that distance tends to zero.

All in all, the entire post that you linked to only plays with the definition of a derivative, the fact that constant additive terms vanish under differentiation, and the fundamental theorem of calculus (FTC).

1. Do not use the same name for the integration variable anywhere else.
2. Do not write $dx$ or $\delta x$ to abbreviate $\lim_{\Delta x \to 0}\dfrac{1}{\Delta x}$ or similar terms. $dx$ has a well-defined meaning as integration variable $\int\, dx$ as well as differentiation variable $\dfrac{d}{dx}$ and another one in differential geometry. It is highly misleading if it is used otherwise, e.g. as integration limits.
3. Forget everything that has been said here about non-standard analysis. This is not the subject here.

 

[JohnnyGui]

See above. The poster said (with a lower bound a=x)
∫xx+ΔXf(t)dt=G(x+Δx)−G(x)
and set h=Δx but this is neither dx nor δx. It is a real number, a distance on the x-axis. We can now consider what happens if that distance h tends to zero, but that doesn't allow us to write dx in the upper bound. That's more than confusing. And dt is doubly wrong.

@fresh_42

Thanks, but I have no idea where you're seeing $\Delta x$ as an upper limit in the source. He is clearly stating $\Delta t$ in the upper limit and says:
$$G(x+\Delta t)-G(x)=\int_x^{x+\Delta t}f(t)dt$$
He is constantly using $\Delta t$ in his expressions afterwards. He might as well be considering $h=\Delta t$ based on his given expressions.

 

[fresh_42]

Not in your link.

https://www.quora.com/What-does-it-...ariable-in-the-upper-bound/answer/Mike-Wilkes

 

[JohnnyGui]

Not in your link.

https://www.quora.com/What-does-it-...ariable-in-the-upper-bound/answer/Mike-Wilkes


View attachment 339955

That is literally a $x+ \Delta t$, not $x + \Delta x$.

 

[fresh_42]

I see that he used $\Delta t$ which is sloppy and not a good idea. Nevertheless, it is not $dt.$

 

[JohnnyGui]

I see that he used $\Delta t$ which is sloppy and not a good idea. Nevertheless, it is not $dt.$

Please look at the source carefully because you said I didn't understand the earlier posts and stated the source completely wrong. I did understand it which is why the source's $\Delta t$ actually confused me.

 

[fresh_42]

Please look at the source carefully because you said I didn't understand the earlier posts and stated the source completely wrong. I did understand it which is why the source's $\Delta t$ actually confused me.

Sorry.

You insisted on writing $x+ dt$ which is not there. Yes, you are right, the mixture of $t$ and $x$ is very confusing. $t$ in $\Delta t$ of the upper bound is technically not the same $t$ as in $\int f(t)\,dt$ and that caused the confusion. This is why I consequently only used $t$ as the integration (dummy) variable and $x$ as the function variable to be clear in my explanation. $\Delta t$ should have been $\Delta x$ and $t$ on the axis should have been $x$ in my opinion. But then you could ask what the difference between $\int f(t)\,dt$ and $\int f(x)\,dx$ would be, which the answer is none. It is in general problematic to overload a letter with different meanings. Imagine we would use $d$ as function variable and limits:
$$
\int_d^{d+\Delta d}f(d)\,dd
$$
makes perfect sense, but is a notational catastrophe!

Nevertheless, the upper limit is a fixed distance on the $x$ or $t$ axis. It cannot be taken as $dx$ or $dt$ since these have other meanings.

 

[PeroK]

@fresh_42

Thanks, but I have no idea where you're seeing $\Delta x$ as an upper limit in the source. He is clearly stating $\Delta t$ in the upper limit and says:
$$G(x+\Delta t)-G(x)=\int_x^{x+\Delta t}f(t)dt$$
He is constantly using $\Delta t$ in his expressions afterwards. He might as well be considering $h=\Delta t$ based on his given expressions.

The source is wrong/sloppy to use $t$ as a variable outside the integrand. It would be logical to use $\Delta x$ in this case.

I suspect you will find several equally sloppy sources.

 

[JohnnyGui]

You insisted on writing x+dt which is not there.

Yes but not insisted though. I only said it in 1 post because the source ultimately used $\Delta t \rightarrow 0$ for his conclusions, which I thought is the same as $dt$. After your correction about this I kept using $\Delta t$ instead.

@PeroK @fresh_42

Thank you for the further explanations.
So judging from all this I can say that it is wrong/sloppy to put a piece $\Delta$ of the integrand's variable in the integral's limits, if the limits contain another variable.

 

[fresh_42]

So judging from all this I can say that it is wrong/sloppy to put a piece $\Delta$ of the integrand's variable in the integral's limits, if the limits contain another variable.

It is questionable to use the integration variable $t$ indicated by $dt$ under the integral anywhere else.

The integral reads $\int_a^b f(t)\,dt =\int_{t=a}^{t=b} f(t)\,dt.$ This resulted in the equation $t=x+\Delta t$ for the upper bound as used in your linked article. It is disturbing to have the same variable $t$ in one equation but with two meanings! $t=x+\Delta x$ would have been the better choice.

It is further problematic to substitute $\Delta x$ by $\delta x$ or $dx.$ They have different meanings, even in case we consider $\Delta x \to 0.$ The understanding of $dx$ in their various contexts is difficult enough even without adding another context.

 

[JohnnyGui]

It is questionable to use the integration variable $t$ indicated by $dt$ under the integral anywhere else.

The integral reads $\int_a^b f(t)\,dt =\int_{t=a}^{t=b} f(t)\,dt.$ This resulted in the equation $t=x+\Delta t$ for the upper bound as used in your linked article. It is disturbing to have the same variable $t$ in one equation but with two meanings! $t=x+\Delta x$ would have been the better choice.

It is further problematic to substitute $\Delta x$ by $\delta x$ or $dx.$ They have different meanings, even in case we consider $\Delta x \to 0.$ The understanding of $dx$ in their various contexts is difficult enough even without adding another context.

Thank you.

How about the upper limit of the equation in pbuk's post of this thread if we substitute $\delta x$ by $\Delta x$ giving:
$$\lim_{\Delta x \to 0} \int_a^{a+\Delta x} f(x) dx \approx \lim_{\Delta x \to 0} f(a) \Delta x$$
I'd expect you won't agree with this upper limit, even if $a$ is considered a value and not a variable? After all, $x=a+\Delta x$ would also show different meanings for $x$.

 

[fresh_42]

Thank you.

How about the upper limit of the equation in pbuk's post of this thread if we substitute $\delta x$ by $\Delta x$ giving:
$$\lim_{\Delta x \to 0} \int_a^{a+\Delta x} f(x) dx \approx \lim_{\Delta x \to 0} f(a) \Delta x$$
I'd expect you won't agree with this upper limit, even if $a$ is considered a value and not a variable? After all, $x=a+\Delta x$ would also show different meanings for $x$.

It breaches my first rule: never use the integration variable elsewhere. I would write it as
$$\lim_{\Delta a \to 0} \int_a^{a+\Delta a} f(x) dx \approx \lim_{\Delta a \to 0} f(a) \Delta a$$
Why use $x$ if we already have $a$ for values on the $x$-axis? If you don't like the same $a$ in $a+\Delta a$ then $\Delta a = h$ is your letter of choice.

However, this rule is for clarity. As mentioned earlier $\int_d^{d+\Delta d}f(d)dd$ is possible to write. We use the same letter for quite a couple of different meanings and its position provides context and meaning. You would certainly agree that such a formula is a nightmare rather than an integral. You could do it, but I strongly recommend not. Same with the $x$ in its double role as an integration (dummy) variable and as an upper limit. You can do it, but it causes threads like this one with meanwhile 30+ posts.

More important is the idea behind the formula. Assume that $F$ is an anti-derivative of $f,$ i.e. $F'=f.$ Let further be $F(x)=F(a)+F'(a)(x-a)+ O((x-a)^2)$ the Taylor series of $F.$ Then
\begin{align*}
\lim_{\Delta a \to 0} \int_a^{a+\Delta a} f(x) dx&=\lim_{\Delta a \to 0}( F(a+\Delta a)-F(a))\\
&=\lim_{\Delta a \to 0}(F(a)+F'(a)\cdot(a+\Delta a -a)+ O((a+\Delta a -a)^2) - F(a))\\
&=\lim_{\Delta a \to 0}(f(a)\cdot \Delta a +O((\Delta a)^2))\\
&\approx\lim_{\Delta a \to 0}f(a)\cdot \Delta a
\end{align*}
Imagine if I had used $\Delta x$ instead of $\Delta a.$ How would you know the difference to the $x$ I used as a variable in the Taylor series? It would open up another discussion about naming objects in mathematics. I like the motto: different meaning requires different letter.