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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.

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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

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

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  • $\begingroup$ Well no, I was not aware of that rule. How do you get it? $\endgroup$ – Bruno Diaz Nov 12 '13 at 14:22
  • $\begingroup$ You might also find this instructive. For the Rule used, see: mathalino.com/reviewer/advance-engineering-mathematics/… $\endgroup$ – Amzoti Nov 12 '13 at 14:25
  • $\begingroup$ Thanks, very good reference. $\endgroup$ – Bruno Diaz Nov 12 '13 at 14:28
  • $\begingroup$ Needs another and another + today. $\endgroup$ – mrs Nov 13 '13 at 8:44
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$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\dd}{{\rm d}}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\ic}{{\rm i}}% \newcommand{\imp}{\Longrightarrow}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert #1 \right\vert}% \newcommand{\yy}{\Longleftrightarrow}$ \begin{align} &\color{#0000ff}{\large\int_{0}^{\infty}\expo{-st}\bracks{t\cos\pars{t}}\,\dd t} = \Re\int_{0}^{\infty}t\expo{\pars{-s + \ic}t}\,\dd t \\[3mm]&= \Re\braces{% \left.t\,{\expo{\pars{-s + \ic}t} \over -s + \ic}\right\vert_{0}^{\infty} - \int_{0}^{\infty}{\expo{\pars{-s + \ic}t} \over -s + \ic}\,\dd t} = \Re\braces{% \left.-\,{1 \over -s + i} {\expo{\pars{-s + \ic}t} \over - s + i}\right\vert_{0}^{\infty}} = \Re\bracks{{1 \over \pars{-s + \ic}^{2}}} \\[3mm]&= \Re\bracks{{\pars{-s - \ic}^{2} \over \pars{s^{2} + 1}^{2}}} = \Re\bracks{s^{2} + 2s\,\ic - 1 \over \pars{s^{2} + 1}^{2}} = \color{#0000ff}{\large{s^{2} - 1 \over \pars{s^{2} + 1}^{2}}} \end{align}

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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 $n=1$

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

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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.

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