Checking if an integral is less than infinite

[atrus_ovis]

Homework Statement

show if $$\int$$$$\stackrel{+\infty}{-\infty}$$|e^-3tsin(t)u(t)|dt < $$\infty$$
where u(t) is the unit step function,
u(t)=1 for t>=0
u(t) = 0 otherwise

The Attempt at a Solution

This is my solution, is it correct / sufficient?

-The integral's bounds can be set from 0 to $$\infty$$, since u(t), and thus the whole quantity to be integrated, is 0 for t<0.
-e^-3t approaches zero, as t approaches $$\infty$$
-|sin(t)| is periodic,with upper & lower bounds 1 and 0 respectively.
Thus the whole expression ->0 as t->$$\infty$$, and the integral is less than $$\infty$$

 

[Dick]

f(x)=1/x goes to 0 as x->infinity. But the integral of that doesn't converge. You need to show more than just that the function goes to 0.

 

[ystael]

No, this is not a correct proof.

The reduction to $$\int_0^\infty e^{-3t}\sin t\,dt$$ is correct. However, in order to prove that $$\int_0^\infty f(t)\,dt < \infty$$, it is not sufficient to show that $$f(t) \to 0$$ as $$t \to \infty$$. (For instance, let $$f(t) = 1$$ for $$0 \leq t \leq 1$$, $$f(t) = 1/t$$ for $$t > 1$$.)

 

[atrus_ovis]

Mm, yeah you're right..Any ideas,general or specific?
The limit of the expression when t->oo is zero btw.

 

[Dick]

Mm, yeah you're right..Any ideas,general or specific?
The limit of the expression when t->oo is zero btw.

Use a comparison test to show the integral exists. Find a function such that |e^(-3t)*sin(t)|<f(t) where you can show the integral of f(t) exists.

 

[atrus_ovis]

Use a comparison test to show the integral exists. Find a function such that |e^(-3t)*sin(t)|<f(t) where you can show the integral of f(t) exists.

But since the original function approaches 0 asymptotically (doesn't it?), won't i have the exact same problem with the comparison function?

 

[Dick]

But since the original function approaches 0 asymptotically (doesn't it?), won't i have the exact same problem with the comparison function?

Pick f(t) so you can integrate it. Being able to integrate it shows it exists. Then if |g(t)|<=f(t) that shows the integral of g(t) exists.

 

[atrus_ovis]

:S
But if i can integrate it, it won't approach asymptotically to zero -> it won't be greater than the original function.
Note that i have to find 'if' it is less than infinity, not to prove it.

 

[Dick]

:S
But if i can integrate it, it won't approach asymptotically to zero -> it won't be greater than the original function.
Note that i have to find 'if' it is less than infinity, not to prove it.

I'm not quite sure what you are getting at here. Is the integral of e^(-3t) from 0 to infinity less than infinity?

 

[atrus_ovis]

I do not know.
e^(-3t) approaches zero, the integral though, i can't tell..

 

[Dick]

I do not know.
e^(-3t) approaches zero, the integral though, i can't tell..

Do the integration! It's an easy antiderivative. Integrate e^(-3t) from 0 to M and then take the limit as M->infinity. What do you get?

 

[atrus_ovis]

Yes, i digged that up via google just now.
Let me work through it.
(it's Integral( e^(-3t)sin(t) dt) btw )

 

[Dick]

Yes, i digged that up via google just now.
Let me work through it.
(it's Integral( e^(-3t)sin(t) dt) btw )

Actually it's not e^(-3t)sin(t). And that's a little bit hard to do. It's |e^(-3t)*sin(t)|. And that one is REALLY hard to do. I'm just asking you to just do the simple one, e^(-3t). Then argue that if that one exists, then the other one exists by a comparison test.

 

[atrus_ovis]

Awh, damn then.I started doing the e^(-3t)sin(t) through uv substitution.
Well:
Integral from 0 to$$\infty$$ of e^(-3t), can be rewritten as the integral from 0 to k, where k->$$\infty$$ of e^(-3t)
I(e^(-3t)) is of course -(e^(-3t) )/3) | from o to k, and plugging in we get
1/3 * (-e^(-3k)+e⁰), which out of the blue, equals 1/3, since e^-$$\infty$$ equals 0 i suppose!

i totally did not see this coming

if the above was right, the problem here is how do we fit in this result to the starting integral

 

[Dick]

Awh, damn then.I started doing the e^(-3t)sin(t) through uv substitution.
Well:
Integral from 0 to$$\infty$$ of e^(-3t), can be rewritten as the integral from 0 to k, where k->$$\infty$$ of e^(-3t)
I(e^(-3t)) is of course -(e^(-3t) )/3) | from o to k, and plugging in we get
1/3 * (-e^(-3k)+e⁰), which out of the blue, equals 1/3, since e^-$$\infty$$ equals 0 i suppose!

i totally did not see this coming

if the above was right, the problem here is how do we fit in this result to the starting integral

|e^(-3t)*sin(t)|<=|e^(-3t)|, right? If the integral of e^(-3t) equal to 1/3, can't you say that the integral of |e^(-3t)*sin(t)| is less than or equal to 1/3?

 

[atrus_ovis]

Yes you can!
Thank you so much Dick.
I did learn something today.