ganeshie8
  • ganeshie8
show that \[\int\limits_0^{\infty}\cos x\,dx = 0\]
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
dan815
  • dan815
i dont think mathematicians would like this very much
Empty
  • Empty
Well 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
  • ganeshie8
Ahh, 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..

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

Empty
  • Empty
I 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
  • Empty
Well 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
  • ganeshie8
yeah, 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
  • ganeshie8
if 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
  • Empty
Well, I suppose by this logic let's just plug in k=-1 and this integral equals -1/2
ganeshie8
  • ganeshie8
\(k\gt 0\)
ganeshie8
  • ganeshie8
http://www.wolframalpha.com/input/?i=%5Cint%5Climits_0%5E%7B%5Cinfty%7D+e%5E%28-kx%29cos%28x%29
Empty
  • Empty
Hmmm this is interesting, but I guess k>0 means that something about this limit isn't quite right
ganeshie8
  • ganeshie8
yeah
Empty
  • Empty
I 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
  • anonymous
I mean, it can't be shown because it isn't true - that integral doesn't converge.
anonymous
  • anonymous
If 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
  • ganeshie8
Right, and aren't we concerned only with the one side limit ? why do we care if it is continuous or not ?
anonymous
  • anonymous
It's less that the function isn't continuous, and more that it isn't defined.
ganeshie8
  • ganeshie8
\(\lim\limits_{k\rightarrow 0+} I(k) = 0 \implies \int\limits_{0}^{\infty} \cos(x)\,dx = 0\) we don't really need the function to be continuous at \(x=0\) right
anonymous
  • anonymous
No, 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
  • anonymous
There'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
  • IrishBoy123
\[\int\limits_0^{\infty}\cos x\,dx = \mathcal {Re} \, \mathcal{L} \{1\}_{s=-i}=0\] or something like that :-)
anonymous
  • anonymous
the 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 well-behaved on subsets of its domain
anonymous
  • anonymous
well, not *any* possible real value, but a continuum of values I meant to say
anonymous
  • anonymous
take \(\sin x\) 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 well-behaved 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
  • anonymous
the 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
  • anonymous
the 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
  • anonymous
Came here to say something about Cauchy PV, glad to see it mentioned above
anonymous
  • anonymous
(It wouldn't have come out so eloquently)

Looking for something else?

Not the answer you are looking for? Search for more explanations.