# Where do you write your dx'es?

Like ∫(...)dx or ∫dx(...) ?

Just wondering ;).

lisab
Staff Emeritus
Gold Member
The first one.

Same here

I usually write it after. It makes sense in my head. "Integrate (function) with respect to (variable)".

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

∫dxf(x) has only one reasonable meaning that is (x+c)f(x). I am however using this unreasonable notation extensively when doing Qm since it is much better to be consistent with the notation you usually find in books.

collinsmark
Homework Helper
Gold Member
Usually this one:
$$\int_a^b \frac{1}{x^2} dx,$$
but occasionally I might write it like this:
$$\int_a^b \frac{dx}{x^2},$$
although this is right out
$$\int_a^b dx\frac{1}{x^2},$$
unless I actually mean for the $\frac{1}{x^2}$ not to be integrated.

lisab
Staff Emeritus
Gold Member
Usually this one:
$$\int_a^b \frac{1}{x^2} dx,$$
but occasionally I might write it like this:
$$\int_a^b \frac{dx}{x^2},$$
although this is right out
$$\int_a^b dx\frac{1}{x^2},$$
unless I actually mean for the $\frac{1}{x^2}$ not to be integrated.

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

∫dxf(x) has only one reasonable meaning that is (x+c)f(x). I am however using this unreasonable notation extensively when doing Qm since it is much better to be consistent with the notation you usually find in books.

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.

$$\int_b^a\frac{1}{log(1)} dx$$

collinsmark
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Gold Member
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
$$\mathcal{O}\left\{ \frac{1}{x^2} \right\}$$
and it's generally understood that the above notation is not necessarily equal to
$$\mathcal{O}\{1\}\frac{1}{x^2}$$
or
$$\frac{1}{x^2}\mathcal{O}\{1\}$$

Of course integrals are a little special, since there truly is a multiplication involved by the dx. But even so, if it's okay to scramble things up, would it also be appropriate to say that
$$\int_a^b \frac{1}{x^2}dx$$
is the same thing as
$$\frac{1}{x^2} \int_a^b dx$$
or worse yet,
$$dx \frac{1}{x^2} \int_a^b \ ?$$
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 dx is my point.

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I like to write it as $\int dx$ to emphasize that it can be viewed as an operator that maps a function into a subspace or to the complex plane for a definite integral. Although, you also define scalar products and norms using integrals, so it might not always work to think of it like that.

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.
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".
∫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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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
$$\mathcal{O}\left\{ \frac{1}{x^2} \right\}$$
and it's generally understood that the above notation is not necessarily equal to
$$\mathcal{O}\{1\}\frac{1}{x^2}$$
or
$$\frac{1}{x^2}\mathcal{O}\{1\}$$

Of course integrals are a little special, since there truly is a multiplication involved by the dx. But even so, if it's okay to scramble things up, would it also be appropriate to say that
$$\int_a^b \frac{1}{x^2}dx$$
is the same thing as
$$\frac{1}{x^2} \int_a^b dx$$
or worse yet,
$$dx \frac{1}{x^2} \int_a^b \ ?$$
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 dx is 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.

collinsmark
Homework Helper
Gold Member
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.
Okay, fair enough. It just looks wrong though.

But I do concede that there have been certain instances where I have put stuff to the right of the dx myself, at least in interim steps. So, even if it doesn't look right, I'll agree it might happen from time to time. I also like to keep my unit vectors to the right, so sometimes there's a conflict (and keeping the unit vector to the right usually wins out for me).

Something like calculating the electric field of a line charge,

$$\int_{\frac{-L}{2}}^{\frac{L}{2}} \frac{1}{4 \pi \varepsilon_0} \frac{\lambda}{(x^2 + z^2)}dx \ \hat{r}$$
Then realizing that
$$\hat r = \frac{x}{\sqrt{x^2 + z^2}} \hat x + \frac{z}{{\sqrt{x^2 + z^2}}} \hat z$$
giving me
$$\int_{\frac{-L}{2}}^{\frac{L}{2}} \frac{1}{4 \pi \varepsilon_0} \frac{\lambda}{(x^2 + z^2)}dx \frac{z}{{\sqrt{x^2 + z^2}}} \hat z$$
before simplifying to
$$\frac{\lambda z}{4 \pi \varepsilon_0 } \int_{\frac{-L}{2}}^{\frac{L}{2}} \frac{dx}{(x^2 + z^2)^{\frac{3}{2}}} \hat z$$

So yeah, I guess it happens. But still, I try to keep the dx to the right if possible. It just looks better to me.

$$dx \frac{1}{x^2} \int_a^b \ ?$$

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.

atyy
It depends on the length of the integrand. For short integrands dx goes at the end, for long integrands it goes at the start.

jhae2.718
Gold Member
I'm going to be pedantic and make the claim that the "d" should be typeset in Roman font; ##\mathrm{d}x## is a differential; ##dx## is the product of scalars ##d## and ##x##.

collinsmark
Homework Helper
Gold Member
I'm going to be pedantic and make the claim that the "d" should be typeset in Roman font; ##\mathrm{d}x## is a differential; ##dx## is the product of scalars ##d## and ##x##.
You bring up a good point, and I was wondering about that. There was a time where I would un-italicize the d when 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 italicized ds 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?

arildno
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Dearly Missed
I prefer:
$$\int_{f(x)}^{dx}|_{a}^{b}$$ BobG
Homework Helper
You bring up a good point, and I was wondering about that. There was a time where I would un-italicize the d when 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 italicized ds 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 of San Serriffe? The writer doesn't seem to like San Serriffe.

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D H
Staff Emeritus
I'm going to be pedantic and make the claim that the "d" should be typeset in Roman font; ##\mathrm{d}x## is a differential; ##dx## is the product of scalars ##d## and ##x##.
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.

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

lisab
Staff Emeritus
Gold Member
$$\int_b^a\frac{1}{log(1)} dx$$

Rabble-rouser!

I prefer:
$$\int_{f(x)}^{dx}|_{a}^{b}$$ You too!

AlephZero
I can live with either. But what should be done about conventions like $$\int_S f(x) \cdot d\mathbf{A}$$ for $$\int f(x) \cdot \mathbf{n}\, dS$$?
$$\int_S d\mathbf{A} \cdot f(x)$$ would look horrible IMO. IMO that looks like it should mean
$$\left(\int_S d\mathbf{A}\right) \cdot f(x)$$ which is nonsense.