# Math Help - Improper Integral

1. ## Improper Integral

Hi,

Im trying to solve the following, Integrate[x^2*Cos[w*x], x], lower limit = 0, upper limit = infinity,

Im getting (2*x*Cos[w*x])/w^2 + ((-2 + w^2*x^2)*Sin[w*x])/w^3

I need to evaluate this now between the lower limit of 0 and upper limit of infinity, but am having trouble,

any thoughts?

Thanks!!.....

2. Before taking the antiderivative, change the upper limit to "t". Then, take your antiderivative (answer will have t's in it). Perform the normal subtraction for integrals. Finally, take the limit as t approaches infinity.

can you repost using latex code? just to make sure i have it right

3. Here's the LaTeX version:

Originally Posted by citbquinn
Hi,

Im trying to solve the following,
$\displaystyle \int_0^{\infty} x^2~cos(wx) ~ dx$

Im getting
$\displaystyle \frac{2x~cos(wx)}{w^2} + \frac{(-2 + w^2x^2)~sin(wx)}{w^3}$

I need to evaluate this now between the lower limit of 0 and upper limit of infinity, but am having trouble,

any thoughts?

Thanks!!.....
-Dan

4. Thanks Dan that looks good.

citbquinn,

simply evaluate your antiderivative from 0 to t. your answer will have t's in it. then take the limit of that as t approaches infinity. that's your answer

5. Your integral looks good. Now what happens when you take the limit as x goes to infinity?

-Dan

6. evaluating from 0 to t, gives,

$=\displaystyle\lim_{t\to \infty}\dfrac{(2tcos(wt))}{w^2} + \dfrac{((-2+w^2t^2)sinwt)}{w^3}$

so is this pretty much my answer??

Thank you!!

7. Originally Posted by citbquinn
evaluating from 0 to t, gives,

$=\displaystyle\lim_{t\to \infty}\dfrac{(2tcos(wt))}{w^2} + \dfrac{((-2+w^2t^2)sinwt)}{w^3}$

so is this pretty much my answer??

Thank you!!
Now you actually have to evaluate the limit ><

8. Originally Posted by Prove It
Now you actually have to evaluate the limit ><
Yes,

this is where my problem is, because coswt for example is periodic as t-->inf,
so there is no limiting value??

9. Originally Posted by citbquinn
Yes,

this is where my problem is, because coswt for example is periodic as t-->inf,
so there is no limiting value??
Since $-1 \leq cos(wt) \leq 1$ what can you say about the values of $t \cdot cos(wt)$?

-Dan

10. Originally Posted by topsquark
Since $-1 \leq cos(wt) \leq 1$ what can you say about the values of $t \cdot cos(wt)$?

-Dan
yes....??

$-t \leq cos(wt) \leq t$

11. Originally Posted by citbquinn
yes....??

$-t \leq cos(wt) \leq t$
Let me be more specific. You want
$\displaystyle \lim_{t \to \infty} t \cdot cos(wt) = \left ( \lim_{t \to \infty} t \right ) \cdot \left ( \lim_{t \to \infty} cos(wt) \right )$

We know that $\displaystyle -1 \leq cos(wt) \leq 1$ so cos(wt) is bounded by -1 and 1 so even though the limit is undefined, it is not infinite.

What about the limit of t?

-Dan