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

- ganeshie8

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

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- dan815

i dont think mathematicians would like this very much

- 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

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..

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## More answers

- 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

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

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

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

Well, I suppose by this logic let's just plug in k=-1 and this integral equals -1/2

- ganeshie8

\(k\gt 0\)

- ganeshie8

http://www.wolframalpha.com/input/?i=%5Cint%5Climits_0%5E%7B%5Cinfty%7D+e%5E%28-kx%29cos%28x%29

- Empty

Hmmm this is interesting, but I guess k>0 means that something about this limit isn't quite right

- ganeshie8

yeah

- 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

I mean, it can't be shown because it isn't true - that integral doesn't converge.

- 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

Right, and aren't we concerned only with the one side limit ?
why do we care if it is continuous or not ?

- anonymous

It's less that the function isn't continuous, and more that it isn't defined.

- 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

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

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

\[\int\limits_0^{\infty}\cos x\,dx = \mathcal {Re} \, \mathcal{L} \{1\}_{s=-i}=0\]
or something like that :-)

- 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

well, not *any* possible real value, but a continuum of values I meant to say

- 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

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

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

Came here to say something about Cauchy PV, glad to see it mentioned above

- anonymous

(It wouldn't have come out so eloquently)

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