# Convergence of an integral.

1. Jun 13, 2006

### MathematicalPhysicist

i need to verify my work, i need to evalute the next integral, and decide if it converges or diverges:
$$\int_{1}^{\infty}cos(\frac{2}{x})dx$$

here what i did:
2/t=x (-2/t^2)dt=dx
$$\int_{1}^{\infty}-2cos(t)/t^2dt$$

cos(t)>=-1
-2cos(t)/t^2<=2/t^2
the integral of 1/t^2 between 1 and infinity converges to 1, and therefore also the orginal integral with cosine converges.

is this line of reasoning correct?

2. Jun 13, 2006

### shmoe

When you substituted 2/x=t, you didn't change the endpoints.

You might want to take a moment to think about what the graph of cos(2/x) looks like on [1,infinity)

3. Jun 13, 2006

### MathematicalPhysicist

ok i understand it diverges.
so i get:
$$\int_{2}^{0}\frac{-2cos(t)}{t^2}dt=-\int_{0}^{2}\frac{-2cos(t)}{t^2}dt$$

when
-2cos(t)/t^2>=-2/t^2

and -2/t^2 diverges on those limits.

4. Jun 13, 2006

### shmoe

-2cos(t)/t^2>=-2/t^2 won't help show that diverges. You're bounding it from the wrong side, your bound is 0<=2cos(t)/t^2<=2/t^2, on [0,1] say. Bounding something from above by something divergent tells you nothing.

You can fix this, but why bother with a change of variables at all? You know what cos(2/x) is doing as x->infinity right?

5. Jun 13, 2006

### MathematicalPhysicist

lim cos(2/x)=1
2/x->0
but how do i find the integral of cos(2/x)?

6. Jun 13, 2006

### shmoe

What do you mean "find the integral"? I thought you already agreed it diverged?

7. Jun 13, 2006

### MathematicalPhysicist

then how do i prove it?

8. Jun 13, 2006

### shmoe

cos(2/x)->1 as x->infinity, so the integral over [1,infinity) diverges.

You've surely got a result to this effect, if not just show you can make the integral over [1,N] arbitrarily large by taking N large enough.

9. Jun 14, 2006

### MathematicalPhysicist

but surely the integral of cos(2/x) is different than cos(2/x), so how the limit of cos(2/x) helps us to prove the integral of cos(2/x) diverges?

10. Jun 14, 2006

### siddharth

Here's one way of looking at it. Consider the function f(x) = cos(2/x). Now as $x \rightarrow \infty$, f(x) tends to 1. So, when you calculate the area under the graph from 1 to infinity, can you see that the area blows up to infinity?

Last edited: Jun 14, 2006
11. Jun 14, 2006

### MathematicalPhysicist

i uderstand that it diverges but how do i prove it rigourosly?

12. Jun 14, 2006

### arildno

In order to prove this rigorously, use some version of the simple ordering property of the integral:
If the integral of a function f (strictly greater than your own integrand) diverges, then your integral diverges as well ("bigger integrand means bigger integral").

13. Jun 14, 2006

### Hammie

Why not expand it out as an infinite series, then integrate?
It can be shown that the first term diverges, and the remainder converges.

Last edited: Jun 14, 2006
14. Jun 14, 2006

### StatusX

Another way would be to write:

$$\int_1^\infty \cos(2/x)dx = \sum_{n=1}^\infty \int_n^{n+1} \cos(2/x)dx$$

And then use what you know about the convergence of infinite series.

15. Jun 14, 2006

### shmoe

The limit of cos(2/x) is 1, so there is an M where cos(2/x)>1/2 if x>M. How big can you then say the integral of cos(2/x) is on the interval [M,N]?

16. Jun 14, 2006

### VietDao29

Ok, what does it mean if someone says that:
$$\lim_{x \rightarrow \infty} \cos \left( \frac{2}{x} \right) = 1$$?
It means that, for every $$\varepsilon > 0$$, there exists a $$\delta > 0$$, such that $$\forall x > \delta \ : \ \left| \cos \left( \frac{2}{x} \right) - 1 \right| < \varepsilon$$
As x increases without bound, $$\cos \left( \frac{2}{x} \right) < 1$$, so:
$$\cos \left( \frac{2}{x} \right) - 1 < 0$$, breaking the absolute value, we have:
$$\forall x > \delta \ : \ - \cos \left( \frac{2}{x} \right) + 1 < \varepsilon$$
$$\forall x > \delta \ : \ \cos \left( \frac{2}{x} \right) > 1 - \varepsilon$$
Now, you can choose $$\varepsilon = \frac{1}{2}$$ as shmoe.
So we have:
$$\forall x > \delta \ : \ \cos \left( \frac{2}{x} \right) > 1 - \frac{1}{2} = \frac{1}{2}$$
So we have:
$$\int_{\delta} ^ \infty \cos \left( \frac{2}{x} \right) dx > \int_\delta ^ \infty \frac{dx}{2}$$.
Does the integral $$\int_{\delta} ^ \infty \cos \left( \frac{2}{x} \right) dx$$ converge or diverge? :)
----------------------
Ok, it works just like series, we know that if the series:
$$\sum_{n = \alpha} ^ {\infty} a_n$$ converges then $$\lim_{n \rightarrow \infty} a_n = 0 , \ n \in \mathbb{N}$$.
So if: $$\int_{\alpha} ^ \infty f(x) dx$$ converges, then $$\lim_{x \rightarrow \infty} f(x) = 0$$. :)
Note that the reverse is, however not true.

Last edited: Jun 14, 2006
17. Jun 14, 2006

### arildno

Why bother with FOTC at all???