Question

integral of sin(1/x)dx between 0 to 1 converges, why?

Answer 1

(Improper integral of course) :)

Answer 2

let u=1/x

Answer 3

It does not work with u substitution.. on WolframAlpha it's something which I have never seen :(

Answer 4

oh yeah, sorry, just realized I messed up :P

Answer 5

The integral from 0 to 1 DOES converge, but I cant understand why.. (any comparasion test or something..?)

Answer 6

An integral is a sum. A sum converges when?

Answer 7

Can only provide an intuitive picture for now.

Answer 8

But a sum works only for Natural numbers.. and where n (the index) starts from? The integral is from 0 to 1, so the sum is from 0 to infinity? [wolfram says the sum diverges. I don't know if in this case you can use sums :X

Answer 9

Again, I can only provide a picture to you. sin will never exceed 1, that I am sure of.

Answer 10

But this is an interesting question. Let me think about it.

Answer 11

Not a great proof of anything, but as a comparative insight, you can notice that\[-1\le\sin\frac1x\le1, x\in\mathcal R\]so the function will never go off to infinity, as I'm sure you already realized. Pictorially the function starts oscillating like crazy as \(x\to0\), but there will always be a curve below the x-axis to cancel the one above in that region, so the integral should be tend to zero around x=0. As you go out towards x=1, you are clearly defined and bounded, so you get some finite answer.

Answer 12

Intuitively you are 100% right, but is there any mathematical method to assure this? Sometime intuition in math does not work :)

Answer 13

I'm trying to conjure up something more concrete, but so far all I have is that...

Answer 14

Related: 1 - 1 + 1 - 1 ... = 0.5
Some people argue that this has no definite value while some agree that it is 0.5.

Answer 15

The problem here is that there is no limit to sine as the input approaches infinite. http://math.stackexchange.com

Answer 16

@oshrinuri , the limits are flawed. Aren't they?

Answer 17

No, it's from 0 to 1. They didn't ask us what is the value of the integral, only whether it converges / diverges..

Answer 18

\[\large \int_0^1 \frac{\sin x}{x}dx= \lim_{R \to 0} \int_R^1 \frac{\sin x}{x}dx= \lim_{R \to 0 } \left. -\frac{\cos x}{x} \right|^1_R - \int_R^1 \frac{\cos x}{x^2}dx \]

Answer 19

\[ \lim_{x \to 0 } \frac{\cos x}{x}=\frac{1}{x \to 0}\sim \infty \]

Answer 20

but it's\[\int_0^1\sin\frac1xdx\]

Answer 21

oh lord, that makes sense then ^^ thank you.

Answer 22

stupid question :'(

Answer 23

can we just say\[-\int_0^1dx\le\int_0^1\sin\frac1xdx\le\int_0^1dx\]?
XD
seems silly, but this would suggest that it converges, right?

Answer 24

another way would be the taylor expansion, studying the radius of convergences and see that it lies within the given bounds of integration.

Answer 25

nevertheless, the above approximation holds true, in fact you can use absolute convergence:
\[\large 0\leq\left| \sin \left( \frac{1}{x} \right) \right| \leq 1 \] and use that absolute convergences implies convergences. Which is what @TuringTest did, holds for approximations for sums in special

Answer 26

Thanks guys! :D