# How to find the Laplace transform of $t\cos{t}$?

I need to find the Laplace transform of $f(t) = t \cos{t}$. I tried using the Taylor series expansion for $\cos{t}$ but I got stuck since the resulting expression is again a series which I could not simplify.

I would like to know if there is an easier method to find this transform without using Taylor series. I don't want the answer, just need to learn how to find it.

Thanks.

We have:

$$\mathcal{L}(\cos(t)) = \dfrac{s}{s^2+1}$$

$$\mathcal{L}(t \cos(t)) = -\dfrac{d}{ds} \left(\dfrac{s}{s^2+1}\right) = \dfrac{s^2-1}{(s^2+1)^2}$$

Are you familiar with the rule I am using?

If

$$\mathcal{L}(f(t)) = F(s)$$

Then

$$\mathcal{L}(t^nf(t)) = (-1)^n\dfrac{d^n}{ds^n}F(s)$$

• Well no, I was not aware of that rule. How do you get it? – Bruno Diaz Nov 12 '13 at 14:22
• You might also find this instructive. For the Rule used, see: mathalino.com/reviewer/advance-engineering-mathematics/… – Amzoti Nov 12 '13 at 14:25
• Thanks, very good reference. – Bruno Diaz Nov 12 '13 at 14:28
• Needs another and another + today. – mrs Nov 13 '13 at 8:44


We know $$L(t^n)=\frac{n!}{s^{n+1}}$$ for integer $n\ge0$

Using Shifting property, $$L\{e^{at}f(t)\}=F(s-a)\text{ where } L\{f(t)\}=F(s)$$

$$\implies L\{e^{at}t^n\}=\frac{n!}{(s-a)^{n+1}}$$

Now, put $a=i=\sqrt{-1}$ to utilize Euler's Formula

$$\implies L\{e^{it}t^n\}=\frac{n!}{(s-i)^{n+1}}=\frac{n!(s+i)^{n+1}}{(s^2+1)^{n+1}}$$

Equate the real parts to find $L\left(\cos t\cdot t^n\right)$

Here $n=1$

$\cal L\{t\cos t\}=-\dfrac{d}{ds}L\{\cos t\}=-\dfrac{d}{ds}\left(\dfrac{s}{s^2+1}\right)=\dfrac{s^2-1}{(s^2+1)^2}$.

If you guys don't mind, I'm going to go ahead and post an additional way to get the solution.

Let $f(t) = t\cos t$. It 's clear that $f(0)=0$.

Now, note that $f^{\prime}(t) = \cos t - t\sin t$ and that $f^{\prime}(0)=1$.

Furthermore, we have that $f^{\prime\prime}(t) = -2\sin t - t\cos t$.

We now make use of the fact that $$\mathcal{L}\{f^{\prime\prime}(t)\} = s^2\mathcal{L}\{f(t)\} - sf(0) -f^{\prime}(0)$$ to see that we have

\begin{aligned} s^2\mathcal{L}\{t\cos t\} - 1 = \mathcal{L}\{-2\sin t - t\cos t\} &\implies (s^2+1)\mathcal{L}\{t\cos t\} = -2\mathcal{L}\{\sin t\}+1\\ &\implies (s^2+1)\mathcal{L}\{t\cos t\} = \frac{s^2-1}{s^2+1} \\ &\implies \phantom{(s^2+1)}\mathcal{L}\{t\cos t\} = \frac{s^2-1}{(s^2+1)^2}\end{aligned} which agrees with what everyone else found.