- #1

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Like ∫(...)dx or ∫dx(...) ?

Just wondering ;).

Just wondering ;).

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- Thread starter PhysicsGente
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- #1

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Like ∫(...)dx or ∫dx(...) ?

Just wondering ;).

Just wondering ;).

- #2

lisab

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The first one.

- #3

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Same here

- #4

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I think it makes more sense before the function though. As just a part of the "integrate" symbol.

- #5

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- #6

collinsmark

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[tex] \int_a^b \frac{1}{x^2} dx,[/tex]

but occasionally I might write it like this:

[tex] \int_a^b \frac{dx}{x^2}, [/tex]

although this is right out

[tex] \int_a^b dx\frac{1}{x^2}, [/tex]

unless I actually mean for the [itex] \frac{1}{x^2} [/itex] not to be integrated.

- #7

lisab

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[tex] \int_a^b \frac{1}{x^2} dx,[/tex]

but occasionally I might write it like this:

[tex] \int_a^b \frac{dx}{x^2}, [/tex]

although this is right out

[tex] \int_a^b dx\frac{1}{x^2}, [/tex]

unless I actually mean for the [itex] \frac{1}{x^2} [/itex] not to be integrated.

Exactly why I don't like it before - it's ambiguous to me.

- #8

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I disagree, placing the dx next to the integral sign should mean the same thing. I don't think we have to necessarily view dx as the "end" or closure of the integral.

- #9

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[tex] \int_b^a\frac{1}{log(1)} dx[/tex]

- #10

collinsmark

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I disagree, placing the dx next to the integral sign should mean the same thing. I don't think we have to necessarily view dx as the "end" or closure of the integral.

Well, the thing is, an integral is an operator. It needs to operate on something.

With lots of other operators, there's different notation involved such as

[tex] \mathcal{O}\left\{ \frac{1}{x^2} \right\} [/tex]

and it's generally understood that the above notation is not necessarily equal to

[tex] \mathcal{O}\{1\}\frac{1}{x^2} [/tex]

or

[tex] \frac{1}{x^2}\mathcal{O}\{1\} [/tex]

Of course integrals are a little special, since there truly is a multiplication involved by the

[tex] \int_a^b \frac{1}{x^2}dx [/tex]

is the same thing as

[tex] \frac{1}{x^2} \int_a^b dx [/tex]

or worse yet,

[tex] dx \frac{1}{x^2} \int_a^b \ ?[/tex]

If the integral sign specifies the left of what must be integrated, what is to specify the right end? That's usually the job of the

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- #11

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- #12

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The way I see it in very non rigorous terms is that you can't chose to not integrate everything that has to do with x in an expression so dx is really "the end of the integral".I disagree, placing the dx next to the integral sign should mean the same thing. I don't think we have to necessarily view dx as the "end" or closure of the integral.

∫dI=∫f(x)dx+ g(x) doesn't mean much. Once you decide to integrate over dx you integrate everything you can't leave any of the xes out .Which when I think about it makes me view the ∫dxf(x) more acceptable somehow.

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- #13

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Well, the thing is, an integral is an operator. It needs to operate on something.

With lots of other operators, there's different notation involved such as

[tex] \mathcal{O}\left\{ \frac{1}{x^2} \right\} [/tex]

and it's generally understood that the above notation is not necessarily equal to

[tex] \mathcal{O}\{1\}\frac{1}{x^2} [/tex]

or

[tex] \frac{1}{x^2}\mathcal{O}\{1\} [/tex]

Of course integrals are a little special, since there truly is a multiplication involved by thedx. But even so, if it's okay to scramble things up, would it also be appropriate to say that

[tex] \int_a^b \frac{1}{x^2}dx [/tex]

is the same thing as

[tex] \frac{1}{x^2} \int_a^b dx [/tex]

or worse yet,

[tex] dx \frac{1}{x^2} \int_a^b \ ?[/tex]

If the integral sign specifies the left of what must be integrated, what is to specify the right end? That's usually the job of thedxis my point.

No one said anything about moving the integration symbol itself, that is completely unrelated to what we are talking about. The integral sign is not a multiplicand; dx is, you said it yourself.

Look at our notation for the limit. There is no "closing" symbol, it's exactly the same thing. dx is not defined to be the end of an integral. To say that putting dx next to the integral sign is not ambiguous is just silly.

- #14

collinsmark

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Okay, fair enough. It just looks wrong though.No one said anything about moving the integration symbol itself, that is completely unrelated to what we are talking about. The integral sign is not a multiplicand; dx is, you said it yourself.

Look at our notation for the limit. There is no "closing" symbol, it's exactly the same thing. dx is not defined to be the end of an integral. To say that putting dx next to the integral sign is not ambiguous is just silly.

But I do concede that there have been certain instances where I have put stuff to the right of the

Something like calculating the electric field of a line charge,

[tex] \int_{\frac{-L}{2}}^{\frac{L}{2}} \frac{1}{4 \pi \varepsilon_0} \frac{\lambda}{(x^2 + z^2)}dx \ \hat{r} [/tex]

Then realizing that

[tex] \hat r = \frac{x}{\sqrt{x^2 + z^2}} \hat x + \frac{z}{{\sqrt{x^2 + z^2}}} \hat z[/tex]

giving me

[tex] \int_{\frac{-L}{2}}^{\frac{L}{2}} \frac{1}{4 \pi \varepsilon_0} \frac{\lambda}{(x^2 + z^2)}dx \frac{z}{{\sqrt{x^2 + z^2}}} \hat z [/tex]

before simplifying to

[tex] \frac{\lambda z}{4 \pi \varepsilon_0 } \int_{\frac{-L}{2}}^{\frac{L}{2}} \frac{dx}{(x^2 + z^2)^{\frac{3}{2}}} \hat z [/tex]

So yeah, I guess it happens. But still, I try to keep the

- #15

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[tex] dx \frac{1}{x^2} \int_a^b \ ?[/tex]

That made my night. Thanks .

I should have said I use the first one when doing math (not physics related), and the second one for physics.

- #16

atyy

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- #17

jhae2.718

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- #18

collinsmark

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You bring up a good point, and I was wondering about that. There was a time where I would un-italicize the

http://mathworld.wolfram.com/AbelsIntegral.html

And Wolfram isn't the only place. It seems pretty common.

So now I don't know what to think.

So where does one learn about such conventions anyway?

- #19

arildno

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I prefer:

[tex]\int_{f(x)}^{dx}|_{a}^{b}[/tex]

[tex]\int_{f(x)}^{dx}|_{a}^{b}[/tex]

- #20

BobG

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You bring up a good point, and I was wondering about that. There was a time where I would un-italicize thedwhen using it for a differential. But then I stopped, not so much because I was lazy, but rather because many others don't seem to do that. Even Wolfram mathworld uses italicizedds such as here:

http://mathworld.wolfram.com/AbelsIntegral.html

And Wolfram isn't the only place. It seems pretty common.

So now I don't know what to think.

So where does one learn about such conventions anyway?

Here would be one place. The actual standard would probably be a better place.

One part of the article does raise a question, though. Is there a math ban in San Serriffe? Or is there a math boycott

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- #21

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I'm not a fan. It's jarring to the eyes (my eyes, in any case). It just looks ugly. Extremely ugly.

And it is *so* very non-508 compliant. (i.e., Section 508 of the US code, www.section508.gov. That link doesn't work right now, thanks to the government shutdown. For now, here's a wikipedia article: http://en.wikipedia.org/wiki/Section_508_Amendment_to_the_Rehabilitation_Act_of_1973.)

With regard to the original question, to me writing ##\int\! dx\; f(x)## is much more esthetically pleasing to me than is ##\int f(x)\,dx##. My rationale: Integration (anti-differentiation) is an operator, so make integration look like an operator. The operator is of course ##\int\! dx##.

However, because too many are confused by that notation, I do tend to use the archaic ##\int f(x)\,dx## form.

- #22

Astronuc

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http://www.ams.org/journals/tran/1905-006-03/S0002-9947-1905-1500719-0/S0002-9947-1905-1500719-0.pdf

http://www.ams.org/journals/tran/1929-031-01/S0002-9947-1929-1501468-1/S0002-9947-1929-1501468-1.pdf

http://mathworld.wolfram.com/Integrand.html

http://mathworld.wolfram.com/Integral.html

- #23

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A long, long time. That's why I called it "archaic".It seems the convention of placing the differential after the integrand has been around a long time.

- #24

lisab

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[tex] \int_b^a\frac{1}{log(1)} dx[/tex]

Rabble-rouser!

I prefer:

[tex]\int_{f(x)}^{dx}|_{a}^{b}[/tex]

You too!

- #25

AlephZero

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$$\int_S d\mathbf{A} \cdot f(x) $$ would look horrible IMO. IMO that looks like it should mean

$$\left(\int_S d\mathbf{A}\right) \cdot f(x) $$ which is nonsense.

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