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ganeshie8
 one year ago
show that \[\int\limits_0^{\infty}\cos x\,dx = 0\]
ganeshie8
 one year ago
show that \[\int\limits_0^{\infty}\cos x\,dx = 0\]

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dan815
 one year ago
Best ResponseYou've already chosen the best response.2i dont think mathematicians would like this very much

Empty
 one year ago
Best ResponseYou've already chosen the best response.2Well this is one way to do it haha \[\lim_{n \to \infty} \int_0^{2 \pi n} \cos(x) dx = \lim_{n \to \infty} [\sin(2 \pi n)  \sin(0)] = \lim_{n \to \infty} 0 = 0\]

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0Ahh, I think I see why mathematicians don't want it to converge \[ \int\limits_{0}^{\infty} \cos x \, dx \text{ converges } \iff \sum\limits_{n=0}^{\infty} \cos(n) \text{ converges} \] which is absurd..

Empty
 one year ago
Best ResponseYou've already chosen the best response.2I don't know if it's that, I think it's this fake notation I made up where the answer is an interval haha \[\lim_{x \to \infty} \cos(x) = [1,1]\]

Empty
 one year ago
Best ResponseYou've already chosen the best response.2Well maybe they're not so different but I know I can make this limit converge by dividing it by x when I take the limit at least.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0yeah, I tend to believe the answer is an interval.. see anythign wrong with below \(I(k)=\int\limits_0^{\infty}e^{kx}\cos x\, dx = \dfrac{k}{k^2+1}\) so \(I(0^{+}) =\int\limits_0^{\infty}\cos x\, dx = \dfrac{0}{0^2+1} = 0 \)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0if that is fine, how do i convince myself about the integral test (\(\sum \cos(n)\) cannot converge) there must be some fien detail that im missing here....

Empty
 one year ago
Best ResponseYou've already chosen the best response.2Well, I suppose by this logic let's just plug in k=1 and this integral equals 1/2

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0http://www.wolframalpha.com/input/?i=%5Cint%5Climits_0%5E%7B%5Cinfty%7D+e%5E%28kx%29cos%28x%29

Empty
 one year ago
Best ResponseYou've already chosen the best response.2Hmmm this is interesting, but I guess k>0 means that something about this limit isn't quite right

Empty
 one year ago
Best ResponseYou've already chosen the best response.2I think this is kind of like trying to do this integral: \[\int x^{n}dx\] since specifically at n=1 we have a special case, beats me.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I mean, it can't be shown because it isn't true  that integral doesn't converge.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0If you're asking what's wrong with your logic for the part where you defined \(I(k)\), then the answer is that that function is defined exclusively for \(k>0\). You're correct in stating that \[ \lim_{k\rightarrow 0+} I(k) = 0\] but \[ \lim_{k\rightarrow a}I(k) = I(a)\] if and only if \(I(k)\) is continuous at \(a\), which it isn't.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0Right, and aren't we concerned only with the one side limit ? why do we care if it is continuous or not ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0It's less that the function isn't continuous, and more that it isn't defined.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0\(\lim\limits_{k\rightarrow 0+} I(k) = 0 \implies \int\limits_{0}^{\infty} \cos(x)\,dx = 0\) we don't really need the function to be continuous at \(x=0\) right

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0No, that's not true. What you've just written is \[ \lim_{k\rightarrow 0+} I(k) = 0 \implies I(0) = 0\] Which is certainly not true if I(0) isn't even defined.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0There's no escaping the fact that, by the definition of an improper Riemann (or any other type) integral, \(\int_0^\infty \cos(x) dx \) does not converge. No amount of algebraic manipulation can change that.

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1\[\int\limits_0^{\infty}\cos x\,dx = \mathcal {Re} \, \mathcal{L} \{1\}_{s=i}=0\] or something like that :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0the reason this integral cannot be said in good faith to converge is because if we manipulate how we choose to have it integrate on our domain of \((0,\infty)\) we can make it converge to any possible real value or even diverge, whereas for integrable functions we expect the integral to be far more wellbehaved on subsets of its domain

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0well, not *any* possible real value, but a continuum of values I meant to say

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0take \(\sin x\) so the intervals are easier to see and observe: for example, if we assume we can break up \((0,\infty)\) we ought to be able to to reduce our integral to the sum of countably many on disjoint intervals \((2\pi k,2\pi(k+1)),k\in\mathbb{N}\) in which case we get $$\int f=0$$whereas if we instead break up our domain into asymmetric pieces like \((0,\pi)\cup(2\pi,3\pi),\quad (\pi,2\pi),\quad(3\pi,4\pi)\cup(5\pi,6\pi),\quad(2\pi,3\pi),\dots\) then we'd get a sum of terms like \(4,2,4,2,4,2,\dots\) which clearly diverges to \(+\infty\). if we reversed the asymmetry we'd get \(\infty\), and if we kept part of this decomposition of the domain into 'bad' pieces for part of the domain we could various arbitrarily large or small finite sums. in a sense, this is why the improper Riemann integral is much like conditionally convergent series (see: rearrangement theorem)  it is not wellbehaved and it is sensitive to *how* you integrate on certain sets. this is why we like the Lebesgue integral and measure theory where we require in our definition of sigma algebras that measures behave reasonably (and this in turn makes integrals work more nicely as well), which is why we can clearly see that \(\cos x\) is not Lebesgue integrable on \(\mathbb{R}^+\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0the Lebesgue integral generalizes the (proper) Riemann integral in ways that allow integrating certain functions on various sets that are simply not possible with the Riemann integral, although it does not generalize the improper Riemann integral

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0the generalization of the Cauchy principal value, with which one can make explicit in waht sense we say $$\int_0^\infty \cos x\, dx=0=\frac12\text{p.v.}\int_{\infty}^\infty \cos x\, dx$$comes to us using the language of distributions, which are a more involved concept

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Came here to say something about Cauchy PV, glad to see it mentioned above

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0(It wouldn't have come out so eloquently)
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