test for divergence for improper integrals?

Do we have "test for divergence" for improper integrals? Recall the test for divergence for infinite series is that if the limit of a_n doesn't approach 0, then the series of a_n diverges. Do we have the same statement for improper integrals?

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#justcalculus

[yoav]

I like this video, because it is something i already know ( the integral of sin(x^2) from 0 to inf converge) but never noticed that the limit of sin(x^2) is not 0 when x goes to inf . You make us think and learn.thank you

[jelmerterburg]

I interpreted "lim(x→∞) f(x) ≠ 0" as implying that the limit exists in the first place, otherwise the ≠ operator has no meaning. But maybe I've had too much exposure to SQL databases, where NULL == 0 and NULL != 0 both evaluate to NULL :P.

[anshumanagrawal]

The limit "Does not exist" Doesn't mean it's "not equal to 0" It's not equal to anything, it's not a number at all, so at least in my opinion when it is said that the limit is not equal to 0, it is reasonable to assume that the limit exists and the number it converges to is a non zero real number, at most limit is not equal to 0 can include when the limit is plus or minus infinity

[mathmathician]

Test for divergence for series does not fully work to improper integral, but one thing that's definitely true is
that:

If f(x) approaches to inf or a finite value (implying the presence of horizontal asymptote) as x approaches to inf (while being continuous everywhere ine the positive interval), then the improper integral from zero to infinity WILL DEFINITELY DIVERGE TO ±inf.

[mdperpe]

I would say that sin(x²) → 0 as x → ∞ in a distributional sense: for every test function φ we have ∫ sin(x²) φ(x-R) dx → 0 as R → ±∞.

[skylardeslypere]

One explanation I thought of is that the areas under the curve (so the first positive bit, then the first negative, and so on) form an alternating series, and as BPRP said, the areas keep decreasing.

Then, by the alternating series test, that sum converges.

The only thing is, I'm not sure whether or not there's a countable set of areas to add up. If it's uncountable, this argument fails.

[factsheet]

The Dirichlet function, is an easy counter example!

what if we considered lim_(x->inf) f(x) = R\{0} ?
in that case i think we could exclude the case where the limit is not defined and still do a valid comparison test

[lily_littleangel]

Suppose now that the limit 'oscilates' in an interval, but 0 is not in this interval, does that mean that the integral converges?

[Anokosciant]

This is true since the limit must exists to talk about it

[sepdronseptadron]

does the lim as x→∞ of f(x) *exists* and ≠0 mean that the integral of f(x) from x=0 to x=∞ div.?
or does it still not work?

[hopelessdove]

Is the statement true if we add the hypothesis that the limit does exist (and isn't 0 ofc)?

[Mariosergio]

Very good. But, how to calculate that integral: (pi/8)^1/2 ???

[ethancheung]

I am not satisfied with this example. the original sequence version, the limit exists. say if we modified the sequence to Bn = [(-1)^n • n]/(2n+1), then lim_n_to_inf also DNE, but the sum of sequence should be finite. this is clearly not the Test for divergence is designed for. So to properly generalise to indefinite integral version, we should not have the example which limit DNE

[thexoxob]

Haven't watched the video, but it's false. A counterexample is sin(x²). Taking the limit as x goes to inf DNE but its integral from 0 to inf is sqrt(pi/8)