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No, technically:
Possibilities = lim->∞
Some possibilities are completely impossible, like fictional surreal stuffs.
Only if you use it for mastermind engineering to conquer the world, then as you said it, but that’s only half true.
Math is a blank slate tool, can be both good or evil depends on the users.
Possibilities = ∞
The indefinite integral of 1/x is ln|x|+c, yes. Definite integrals are numbers.
Very properly speaking, an improper integral is a limit. In this case,
\int_1^\infty (1/x) dx
is evaluated as
\lim_{a \rightarrow \infty} ( \int_1a (1/x) dx )
=
\lim_{a \rightarrow \infty} ( ln|a| - ln 1 )
=
\lim_{a \rightarrow \infty} ln|a|
=
\infty.
In words, the area under the curve 1/x when x ranges from 1 to any number A is just the natural log of A. If you let A get really big, its log gets really big too, and hence the integral diverges.
@VladimirMacHolzraum
@Techy Pony
@Ponyra
Your comments, by the way, would apply if the lower limit were zero. In which case a double limit would be required:
\int_0\infty (1/x) dx
is evaluated as
\lim_{b \rightarrow 0} ( \int_b^1 (1/x) dx )
\lim_{a \rightarrow \infty} ( \int_1^a (1/x) dx )
=
\lim_{b \rightarrow 0} ( ln 1 - ln |b| )
\lim_{a \rightarrow \infty} ( ln|a| - ln 1 )
=
\lim_{b \rightarrow 0} ( - ln |b| )
\lim_{a \rightarrow \infty} ln|a|
=
\lim_{b \rightarrow \infty} ( - ln |1/b| )
\lim_{a \rightarrow \infty} ln|a|
=
\lim_{b \rightarrow \infty} ln |b|
\lim_{a \rightarrow \infty} ln|a|
=
\infty.
Here, we made the substitution that ln|1/b| = minus ln|b|, so as to avoid having to argue about minus ( minus \infty). (Heh, the booru code wants to make a bunch of minus signs a strikethrough.) This changes the limit since as b goes to 0, 1/b goes to infinity. Then we simply get the positive sum of two positive, diverging terms. However, it could have been done without that step.
Edited
Not outstanding, but better than >>1444444
Also, the integral of 1/x is ln|x|. Which is still infinity
Oh, thanks
An older image was merged with this one.
1/1 + 1/2 + 1/3 + 1/4 + …
goes to infinity, because
the area under the curve from 1 to 2 is less than 1/1…
the area under the curve from 2 to 3 is less than 1/2…
the area under the curve from 3 to 4 is less than 1/3…
and so forth, but it still adds up to something that diverges.
Fucking algebra ruining everything.
Edited
your current filter.This is the result by Wolfram Alpha.
By the way, I’m surprised I could remember LaTeX so well.
it would lead to ln(∞)-ln(-∞), or ln(∞/-∞) , which is undetermined
Ooops, yeah your’e right. However, you’d think it’d be equal to 0 since you’re subtracting identically “sized” infinities. Though when it comes to calc and infinities, things are not usually as they seem.