1. ## infinite integral

is the infinite integral of an oscillating function like sin(x) or cos(x) 0? i know that a definite integral of this type will depend on the limits of integration, but over the interval [0, infinity) 0 seems like the only thing that would make sense but i wanna convince myself a little more strongly than that

2. Originally Posted by da kid
is the infinite integral of an oscillating function like sin(x) or cos(x) 0? i know that a definite integral of this type will depend on the limits of integration, but over the interval [0, infinity) 0 seems like the only thing that would make sense but i wanna convince myself a little more strongly than that
$\int_0^\infty\sin x\,dx$

$=\lim_{b\to\infty}\int_0^b\sin x\,dx$

$=\lim_{b\to\infty}\bigg[-\cos x\bigg]_0^b$

$=\lim_{b\to\infty}(-\cos b+\cos0)=\lim_{b\to\infty}(1-\cos b)$

As $b\to\infty,$ is $\cos b$ approaching a single finite value? Then what can we say about this limit?

3. The integral is undefined,
say for sin(x)

$\int_{0}^{a} \sin x = 1 - \cos{a}$

if we take the limit as cos(a) approaches infinity we see it cant converge, if we add some more to a, the integral would always change value it doesnt settle down to just one value over time (crap explanation I know, but I hope you get the gist of what I'm saying!)

4. Originally Posted by Reckoner
$\int_0^\infty\sin x\,dx$

$=\lim_{b\to\infty}\int_0^b\sin x\,dx$

$=\lim_{b\to\infty}\bigg[-\cos x\bigg]_0^b$

$=\lim_{b\to\infty}(-\cos b+\cos0)=\lim_{b\to\infty}(1-\cos b)$

As $b\to\infty,$ is $\cos b$ approaching a single finite value? Then what can we say about this limit?
i guess the limit just doesn't exist. i mean we can split up the difference and take the limit of 1 as b goes to infinity (which is 1 of course) but the limit of cos(b) as b goes to infinity.....well there just isn't one. so this means that the infinite integral of these types of functions does not exist right? that makes sense too. i guess i was thinking that taking the infinite integral somehow took all of infinity into account simultaneously like when we say that the infinite integral of an unbounded function is infinity even tho over any [a,b] interval the definite integral is finite. there too i see that it would help to remember to think in terms of limits. but still it seems somehow different because there is a symbol we use with a specific meaning to denote this integral but in the case of the oscillating function we just kind of throw our hands up and say "it doesn't exist". i see it and i think i get it but the not existing thing doesn't feel like it gives any problem closure. i have the same mental issue with certain objects just being undefined like division by 0.

hey, if you or anyone else knows of a good book that talks about these kinds of mental issues (the ones i have) let me know! thanks!

5. Originally Posted by da kid
i guess the limit just doesn't exist. i mean we can split up the difference and take the limit of 1 as b goes to infinity (which is 1 of course) but the limit of cos(b) as b goes to infinity.....well there just isn't one. so this means that the infinite integral of these types of functions does not exist right? that makes sense too. i guess i was thinking that taking the infinite integral somehow took all of infinity into account simultaneously like when we say that the infinite integral of an unbounded function is infinity even tho over any [a,b] interval the definite integral is finite. there too i see that it would help to remember to think in terms of limits. but still it seems somehow different because there is a symbol we use with a specific meaning to denote this integral but in the case of the oscillating function we just kind of throw our hands up and say "it doesn't exist". i see it and i think i get it but the not existing thing doesn't feel like it gives any problem closure. i have the same mental issue with certain objects just being undefined like division by 0.
Not every improper integral will converge, just as not all series converge. In this case, you have an "infinite" area above the $x$-axis and an "infinite" area below the $x$-axis, so if you were to try to subtract them you would be doing $\infty-\infty,$ which is not defined.

I am not sure that I can give a clearer explanation. Just remember that when it comes to the infinite, you need to leave your intuitions and preconceived notions behind, because things will often have unexpected results.