What's it mean when people write: [eqn] \int_A f(x) dx

A = [a,b]

A is a set. It can be an interval, but it really can be any measurable set

>>7733761
Thanks, from now on I will write all of my integrals like that.

>>7734788
there's a reason there are so many different ways to write integrals. different forms make more sense in different situations.

when you're simply evaluating from a to b it's super autistic to define the integral over a set where the set is [a,b]

when you're using multiple integrals it's often handy to define the integral over (different people use different letters) R (a region in R2), S (a region in R3), C (a parameterized curve), sigma S (the boundary of a region in R3) etc etc

also, stupid questions thread. or questions that don't deserve their own threads thread.

>>7734813
Does it ever happen that people write:

[eqn] \int_{\mathbb{R}} f(x) dx [/eqn]

Instead of:

[eqn] \int_{-\infty}^{\infty} f(x) dx [/eqn]

>>7735300
Yes.

>>7735300
yup

>>7733749

[math] ∫_Ω dω= ∫ _ { ∂Ω } ω [/math]

>>7733749
The first case would probably be stated as

[math]\int_A f(x) dA[/math]

Like in equations for moments of inertia, where you're specifying differentials of area, rather than a differential w/r/t a particular axis. It's convention.

>>7735300

Either notation is fine as long as whichever being used, is understood unambiguously by its users in context, and I'm sure people have used something like the former form.

But where the real line is concerned, the latter form is "the standard convention", and this seems to hold good among every (discipline's) text involving integration over the real line that I've seen (my own math textbooks, a few higher math papers, an old co-worker's engineering handbook). First-year calc students still have to get competent at considering a function's behavior at every point (and treating two or more limiting cases separately), and the latter form is a bit more suggestive of the details that need to be checked before evaluation is to be attempted.

Integration over the complex plane, of course, or (often) contour integration, is more involved.

In case you didn't know, such an integral is known as an /improper integral/, which just means that infinity and/or undefined points are involved within the area of integration, and have to be accounted for. There are ways of getting round non-convergence if you have a simple number of regions (particularly congruent infinite regions), and assigning a value, e.g. Cauchy Principal Value.

https://en.wikipedia.org/wiki/Improper_integral

Try this as an experiment. ask Wolfram Alpha to integrate f(x) = x "over the real line", as we've been discussing. On the one hand, you have two infinite regions which blow up (slowly), and so the integral does not converge. On the other hand, the two regions are intuitively congruent and cancel each other, where the interval of integration is grown symmetrically (and arbitrarily large). Thus the integral has a /principal value/ of zero.

>>7735360
Thank you, my analysis teacher talked about this but he said [math] \int_{\infty}^{\infty} x dx [/math] wasn't defined, because we can take:

[eqn] \int_{\infty}^{\infty} x dx &= \lim_{c\to\infty} \int_{-c}^{2c} x dx \\
&= \lim_{c\to\infty} (\frac{1}{2}(2c)^{2} -\frac{1}{2}c^2 \\
&= \lim_{c\to\infty} 2c^{2}-\frac{1}{2}c^{2} \\
&= +\infty [/eqn]

On the other hand,

[eqn] \lim_{c\to\infty} \int_{-c}^{c} x dx = 0 [/eqn]

>>7735742
Damn it

[eqn] \int_{-\infty}^{\infty} x dx = \lim_{c\to\infty} \int_{-c}^{2c} x dx = \lim_{c\to\infty} (\frac{1}{2}(2c)^{2} -\frac{1}{2}c^2 = \lim_{c\to\infty} 2c^{2}-\frac{1}{2}c^{2} = +\infty [/eqn]

>>7735349
Fuck no. A is the set where you're integrating

>>7735747

This is wrong in two ways. First, the computation is wrong because a 2 snuck in there for some reason, but that's cosmetic. The second important way that this line is wrong, is that it is concluded that the original integral becomes arbitrarily large (tends to positive infinity); this is false. /We cannot even say that about the original integral/. Rather, what you end up with on evaluation is an indeterminate form, being infinity minus infinity, /which can't even be judged one way or the other/. We are reduced to saying that the integral /does not exist/, or a similar unhelpful statement. Its /principal cauchy value/, on the other hand, does exist:

[math] \displaystyle PV \int_{- \infty}^{ \infty} x \;dx = 0 [/math]

https://en.wikipedia.org/wiki/Improper_integral#Convergence_of_the_integral

Generally, the principal value of an integral of an odd function over the real line is 0, as you'd expect, and that of an even function is just double the result in one direction from 0.