Question

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

Answer 1

i dont think mathematicians would like this very much

Answer 2

Well this is one way to do it haha

\[\lim_{n \to \infty} \int_0^{2 \pi n} \cos(x) dx = \lim_{n \to \infty} [\sin(2 \pi n) - \sin(0)] = \lim_{n \to \infty} 0 = 0\]

Answer 3

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..

Answer 4

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

\[\lim_{x \to \infty} \cos(x) = [-1,1]\]

Answer 5

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.

Answer 6

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 \(I(0^{+}) =\int\limits_0^{\infty}\cos x\, dx = \dfrac{0}{0^2+1} = 0 \)

Answer 7

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....

Answer 8

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

Answer 9

\(k\gt 0\)

Answer 10

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

Answer 11

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

Answer 12

yeah

Answer 13

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.

Answer 14

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

Answer 15

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.

Answer 16

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

Answer 17

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

Answer 18

\(\lim\limits_{k\rightarrow 0+} I(k) = 0 \implies \int\limits_{0}^{\infty} \cos(x)\,dx = 0\)

we don't really need the function to be continuous at \(x=0\) right

Answer 19

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.

Answer 20

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.

Answer 21

\[\int\limits_0^{\infty}\cos x\,dx = \mathcal {Re} \, \mathcal{L} \{1\}_{s=-i}=0\]

or something like that :-)

Answer 22

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

Answer 23

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

Answer 24

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 \(\mathbb{R}^+\)

Answer 25

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

Answer 26

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

Answer 27

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

Answer 28

(It wouldn't have come out so eloquently)