impropres integrals

[waxen bison]

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[unborn stag]

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[waxen bison]

The first one

[lilac shell]

, rotate

[lilac shell]

just a thought

is this true?

$$ x\leqslant \exp (\sqrt{\ln x}), \quad x\geqslant 1 $$

[elder pine]

aL

[lilac shell]

if this is true, then the integrand is bounded by 1/x^2 and that converges

@waxen bison

[vernal sun]

Not sure about x >= 1, but for x large enough will suffice

[waxen bison]

Not that one

Lol

[lilac shell]

you said the first one

[waxen bison]

It is sin(x)*sin(1/x)

[lilac shell]

i quote

[vernal sun]

Oh ok

[waxen bison]

I mean n 8 is the only full exercise

[lilac shell]

https://tenor.com/view/crystal-ball-gif-8836058

[vernal sun]

I don't know if there is a simple way of doing the first one

I think you can fall back to series if you cut the integral into chunks on intervals of size pi

[lilac shell]

taylor series maybs

[vernal sun]

and then use a convergence test

[lilac shell]

maybe Abel test

[waxen bison]

Hmm

[vernal sun]

in the neighborhood of 0 there is no problem with convergence due to continuity

but at infinity it's more problematic

[waxen bison]

I have to start with the absolute value

[tardy yacht]

Intégration by parts and comparisons to the Riemann integral of 1/x^2 no ?

Like set du=sin(x) u=-cos(x) v=sin(1/x) dv =-1/x^2 * cos(1/x)

[vernal sun]

Yeah you know what, I think rotor is right on that one

[waxen bison]

Nope it's divergent

[vernal sun]

because the integral of sin(x)/x converges

[waxen bison]

So it's not on the good side of the comparison

[lilac shell]

sin (1/x) ~ 1/x as x gets large

might be helpful

[vernal sun]

Yeah though unfortunately we can't really use equivalents due to the flips in signs

[waxen bison]

I mean is there a trig identity I can use ?

[tardy yacht]

Asymptotic Comparaison only works if the function inside has a constant

Sign

[lilac shell]

analyse series form then maybe

the integrand is continuous around 0 so

[vernal sun]

I think a series + AST is fine

[lilac shell]

$$ \sum _{k=1}^\infty \sin k \sin \frac{1}{k} $$

[elder pine]

aL

[lilac shell]

and now use the comparison test

with some suitably picked series

well comparison yields that is sufficient to study behaviour of

$$ \sum _{k=1}^\infty \frac{\sin k}{k} $$

[elder pine]

aL

[lilac shell]

because $$ \lim _{k\to \infty} \frac{\sin k \sin \frac{1}{k}}{\sin k\cdot \frac{1}{k}} \to 1 $$

[elder pine]

aL

[lilac shell]

oops no, it doesn't converge absolutely, but can't say if it diverges yet

[tardy yacht]

If you set x=$\frac{1}{u}$ the integral becomes $\int_{0}^{\infty} \frac{sin(u)sin(\frac{1}{u})}{u^{2}} du$

[elder pine]

Rotor

[lilac shell]

through some fourier stuff i think this actually converges

so the initial integral should also converge

[vernal sun]

Yeah I guess that's far simpler

that being said now you have problems in the neighborhood of 0

so there's that

[tardy yacht]

Yea fourrier analysis works but also The series expansion of -ln(1-x) converges uniformly for |x|<=1 $x \neq 1$ so setting $sin(k)=Im(e^{ik})$ you have the value of the series I think

[elder pine]

Rotor

[tardy yacht]

Not sure though but I think it works

[lilac shell]

yup, that's how i did it

consider the sum exp (...) thing as a geometric series

[vernal sun]

I'm not sure the comparison test actually holds for series which sign is not constant

[lilac shell]

shoot you might be right, lemme double check

[tardy yacht]

I think you need monotony and continuity of the function

[vernal sun]

That being said, for simple convergence, one can consider $a_k = \int_{k\pi}^{(k+1)\pi} \sin(x) \sin(1/x)dx$

[elder pine]

Rion

[lilac shell]

you're right

[vernal sun]

It is not hard to prove that this sequence alternates and that $\lvert a_{k+1} \rvert < \lvert a_k \rvert$, and that it converges to 0

[elder pine]

Rion

[vernal sun]

So one can go straight for the kill with AST

For absolute convergence however I have no idea

[tardy yacht]

For absolute convergence $\frac{|sin(x)|}{x} \sim |sin(\frac{1}{x})sin(x)|)$ for x approaching infinity

[lilac shell]

my approach is applicable for absolute convergence

and it's divergent

[tardy yacht]

I meant equivalent

[lilac shell]

sim

[elder pine]

Rotor

[tardy yacht]

Thx

[plush echo]

@tardy yacht has given 1 rep to @lilac shell

[tardy yacht]

The integral of |sin(x)|/x is divergent

[lilac shell]

yup, we get a harmonic lower bound

[vernal sun]

I got the same yeah

[lilac shell]

this discussion made me think of something that i cant figure out

[waxen bison]

+close