Improper Integral

**citbquinn** · March 15th 2011, 10:01 AM

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!!.....

**divinelogos** · March 15th 2011, 10:57 AM

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

**topsquark** · March 15th 2011, 11:07 AM

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

**divinelogos** · March 15th 2011, 12:51 PM

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

**topsquark** · March 15th 2011, 01:07 PM

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

-Dan

**citbquinn** · March 15th 2011, 03:55 PM

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!!

**Prove It** · March 15th 2011, 05:57 PM

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 ><

**citbquinn** · March 16th 2011, 02:17 AM

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??

**topsquark** · March 16th 2011, 10:30 AM

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

**citbquinn** · March 16th 2011, 12:50 PM

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....??

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

**topsquark** · March 16th 2011, 05:27 PM

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

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