MHB What Is the Limit of $t \cos(wt) e^{-st}$ as $t \to \infty$?

[evinda]

Hi! :)
I have to find the Laplace transform of the function $tcos(wt) , w>0 $ .
That's what I have done so far:
$I=\int_{0}^{\infty}e^{-st}tcos(wt)dt=\int_{0}^{\infty}(\frac{-e^{-st}}{s})'tcos(wt)dt=\left [(\frac{-e^{-st}}{s})tcos(wt) \right ]_{0}^{\infty}-\int_{0}^{\infty}(\frac{-e^{-st}}{s}(cos(wt)-sin(wt))dt $

How can I find the limit,when $t\to \infty$ ?

 

[I like Serena]

Hi! :)
I have to find the Laplace transform of the function $tcos(wt) , w>0 $ .
That's what I have done so far:
$I=\int_{0}^{\infty}e^{-st}tcos(wt)dt=\int_{0}^{\infty}(\frac{-e^{-st}}{s})'tcos(wt)dt=\left [(\frac{-e^{-st}}{s})tcos(wt) \right ]_{0}^{\infty}-\int_{0}^{\infty}(\frac{-e^{-st}}{s}(cos(wt)-sin(wt))dt $

How can I find the limit,when $t\to \infty$ ?

Hmm, let's see... according to the rule for partial integration, we have:
$$\int udv = uv - \int vdu$$
So we get:
\begin{aligned}
I &= \int_0^\infty t\cos(\omega t)\ d\left(\frac{-e^{-st}}{s}\right) \\
&= \left[ t\cos\omega t \cdot \frac{-e^{-st}}{s} \right]_0^\infty - \int \frac{-e^{-st}}{s} d(t\cos(\omega t)) \\
&= \left[ t\cos(\omega t) \cdot \frac{-e^{-st}}{s} \right]_0^\infty - \int_0^\infty \frac{-e^{-st}}{s} (\cos(\omega t) - \omega t \sin(\omega t)) dt
\end{aligned}
When $t \to \infty$ we get:
$$\infty \cdot \cos(\omega \cdot \infty) \cdot \frac{-e^{-s \cdot \infty}}{s}$$
If $\mathfrak{Re}(s) > 0$, the exponential power will approach 0 faster than any of the rest of the expression diverges. So this part of the result is 0.

 

[evinda]

Hmm, let's see... according to the rule for partial integration, we have:
$$\int udv = uv - \int vdu$$
So we get:
\begin{aligned}
I &= \int_0^\infty t\cos(\omega t)\ d\left(\frac{-e^{-st}}{s}\right) \\
&= \left[ t\cos\omega t \cdot \frac{-e^{-st}}{s} \right]_0^\infty - \int \frac{-e^{-st}}{s} d(t\cos(\omega t)) \\
&= \left[ t\cos(\omega t) \cdot \frac{-e^{-st}}{s} \right]_0^\infty - \int_0^\infty \frac{-e^{-st}}{s} (\cos(\omega t) - \omega t \sin(\omega t)) dt
\end{aligned}
When $t \to \infty$ we get:
$$\infty \cdot \cos(\omega \cdot \infty) \cdot \frac{-e^{-s \cdot \infty}}{s}$$
If $\mathfrak{Re}(s) > 0$, the exponential power will approach 0 faster than any of the rest of the expression. So this part of the result is 0.

But how can I prove it?Using the L'Hospital Rule?

 

[I like Serena]

But how can I prove it?Using the L'Hospital Rule?

Yes.
But first consider that:
$$\frac{-t}{se^{st}} \le t\cos(\omega t) \frac{e^{-st}}{s} \le \frac{+t}{se^{st}}$$
Then apply l'Hospital's rule to both the LHS and the RHS.
Apply the squeeze rule after that.

 

[polygamma]

If you want to be cute, notice that

$$\int_{0}^{\infty} t \cos (\omega t) e^{-st} \ dt = \text{Re} \int_{0}^{\infty} t e^{i \omega t}e^{-st} \ dt = \text{Re} \int_{0}^{\infty} t e^{-t(s-iw)} \ dt$$

$$ = \text{Re} \Bigg( - \frac{t}{s-iw} e^{-t(s-i \omega)} \Big|^{t=\infty}_{t=0} + \int_{0}^{\infty} \frac{1}{s-i \omega} e^{-t(s-i \omega)} \ dt \Bigg) = \text{Re} \int_{0}^{\infty} \frac{1}{s-i \omega} e^{-t(s-i \omega)} \ dt $$

$$ = \text{Re} \frac{1}{(s-i \omega)^{2}} \int_{0}^{\infty} e^{-u} \ du = \text{Re} \ \frac{1}{(s-i \omega)^{2}}$$

$$ = \text{Re} \ \frac{1}{(s-i \omega)^{2}} \frac{(s+i \omega)^{2}}{(s+ i \omega)^{2}} = \text{Re} \ \frac{s^{2} - w^{2} + 2 i \omega s}{(s^{2}+ \omega^{2})^{2}} = \frac{s^{2}-\omega^{2}}{(s^{2}+\omega^{2})^{2}}$$And now we also know that

$$\int_{0}^{\infty} t \sin( \omega t) e^{-st} \ dt = \text{Im} \ \frac{s^{2} - w^{2} + 2 i \omega s}{(s^{2}+ \omega^{2})^{2}} = \frac{2 \omega s}{(s^{2}+\omega^{2})^{2}} $$

 

[evinda]

Yes.
But first consider that:
$$\frac{-t}{se^{st}} \le t\cos(\omega t) \frac{e^{-st}}{s} \le \frac{+t}{se^{st}}$$
Then apply l'Hospital's rule to both the LHS and the RHS.
Apply the squeeze rule after that.

Nice!Thank you very much!

- - - Updated - - -

If you want to be cute, notice that

$$\int_{0}^{\infty} t \cos (\omega t) e^{-st} \ dt = \text{Re} \int_{0}^{\infty} t e^{i \omega t}e^{-st} \ dt = \text{Re} \int_{0}^{\infty} t e^{-t(s-iw)} \ dt$$

$$ = \text{Re} \Bigg( - \frac{t}{s-iw} e^{-t(s-i \omega)} \Big|^{t=\infty}_{t=0} + \int_{0}^{\infty} \frac{1}{s-i \omega} e^{-t(s-i \omega)} \ dt \Bigg) = \text{Re} \int_{0}^{\infty} \frac{1}{s-i \omega} e^{-t(s-i \omega)} \ dt $$

$$ = \text{Re} \frac{1}{(s-i \omega)^{2}} \int_{0}^{\infty} e^{-u} \ du = \text{Re} \ \frac{1}{(s-i \omega)^{2}}$$

$$ = \text{Re} \ \frac{1}{(s-i \omega)^{2}} \frac{(s+i \omega)^{2}}{(s+ i \omega)^{2}} = \text{Re} \ \frac{s^{2} - w^{2} + 2 i \omega s}{(s^{2}+ \omega^{2})^{2}} = \frac{s^{2}-\omega^{2}}{(s^{2}+\omega^{2})^{2}}$$And now we also know that

$$\int_{0}^{\infty} t \sin( \omega t) e^{-st} \ dt = \text{Im} \ \frac{s^{2} - w^{2} + 2 i \omega s}{(s^{2}+ \omega^{2})^{2}} = \frac{2 \omega s}{(s^{2}+\omega^{2})^{2}} $$

Great!I found the same result..! Thanks for your answer! :D

 

[alyafey22]

Let

$$F(\omega) = \int_{0}^{\infty} \sin(\omega t) e^{-st} \ dt=\frac{\omega}{s^2+\omega^2}$$

$$F'(\omega) = \int_{0}^{\infty} t \cos(\omega t) e^{-st} \ dt=\frac{s^2-\omega^2}{(s^2+\omega^2)^2}$$

 

[I like Serena]

If you want to be cute, notice that

$$\int_{0}^{\infty} t \cos (\omega t) e^{-st} \ dt = \text{Re} \int_{0}^{\infty} t e^{i \omega t}e^{-st} \ dt$$

Can you support that your result also holds for complex $s$?

 

[evinda]

Can you support that your result also holds for complex $s$?

Is it necessary to do it with $\text{Re and Im}$ ? I haven't done it like that!

 

[I like Serena]

Is it necessary to do it with $\text{Re and Im}$ ? I haven't done it like that!

No, that is not necessary. It is just a cute way to it. :)
And as yet, I am questioning its validity, since it doesn't take complex values of $s$ into account.

 

[polygamma]

I'm going to take the easy way out of this.

The functions $\int_{0}^{\infty} t \cos (\omega t) e^{-st}$ and $ \frac{s^{2}-\omega^{2}}{(s^{2}+\omega^{2})^{2}}$ agree for $s>0$.

The function $\frac{s^{2}-\omega^{2}}{(s^{2}+\omega^{2})^{2}}$ is analytic for $\text{Re}(s) > 0$.

And by Morera's theorem (in combination with Fubini's theorem) $\int_{0}^{\infty} t \cos (\omega t) e^{-st} \ dt$ is analytic for $\text{Re}(s) > 0$ as well.

So by analytic continuation, they must agree for $\text{Re} (s) > 0 $.