Question

Find the laplace transform by intergration
f(t)=tcosh(3t)

Answer 1

`\mathcal L_s\{t\cosh3t\}=\displaystyle\int_0^\infty t\cosh3t e^(-st)\,\mathrm dt`

Integrate by parts, setting


`u_1=t\implies\mathrm du_1=\mathrm dt`

`\mathrm dv_1=\cosh3t e^(-st)\,\mathrm dt\implies v_1=\displaystyle\int\cosh3t e^(-st)\,\mathrm dt`

To evaluate
`v_1`, integrate by parts again, this time setting


`u_2=\cosh3t\implies\mathrm du_2=3\sinh3t\,\mathrm dt`

`\mathrm dv_2=\displaystyle\int e^(-st)\,\mathrm dt\implies v_2=-\frac1se^(-st)`


`\implies\displaystyle\int\cosh3te^(-st)\,\mathrm dt=-\frac1s\cosh3te^(-st)+\frac3s\int \sinh3te^(-st)`

Integrate by parts yet again, with


`u_3=\sinh3t\implies\mathrm du_3=3\cosh3t\,\mathrm dt`

`\mathrm dv_3=e^(-st)\,\mathrm dt\implies v_3=-\frac1se^(-st)`


`\implies\displaystyle\int\cosh3te^(-st)\,\mathrm dt=-\frac1s\cosh3te^(-st)+\frac3s\left(-\frac1s\sinh3te^(-st)+\frac3s\int\cosh3te^(-st)\,\mathrm dt\right)`

`\displaystyle\int\cosh3te^(-st)\,\mathrm dt=-\frac1s\cosh3te^(-st)-\frac3{s^2}\sinh3te^(-st)+\frac9{s^2}\int\cosh3te^(-st)\,\mathrm dt`

`\displaystyle(s^2-9)/(s^2)\int\cosh3te^(-st)\,\mathrm dt=-\frac1s\cosh3te^(-st)-\frac3{s^2}\sinh3te^(-st)`

`\implies\displaystyle\underbrace{\int\cosh3te^(-st)\,\mathrm dt}_(v_1)=-((s\cosh3t+3\sinh3t)e^(-st))/(s^2-9)`

So we have


`\displaystyle\int_0^\infty t\cosh3t e^(-st)\,\mathrm dt=u_1v_1\big|_(t=0)^(t\to\infty)-\int_0^\infty v_1\,\mathrm du_1`

`=\displaystyle-(t(s\cosh3t+3\sinh3t)e^(-st))/(s^2-9)\bigg|_(t=0)^(t\to\infty)-\int_0^\infty \left(-((s\cosh3t+3\sinh3t)e^(-st))/(s^2-9)\right)\,\mathrm dt`

`=\displaystyle\frac1{s^2-9}\int_0^\infty(s\cosh3t+3\sinh3t)e^(-st)\,\mathrm dt`

We already have the antiderivative for the first term:


`\displaystyle\frac s{s^2-9}\int_0^\infty \cosh3te^(-st)\,\mathrm dt=\frac s{s^2-9}\left(-((s\cosh3t+3\sinh3t)e^(-st))/(s^2-9)\right)\bigg|_(t=0)^(t\to\infty)`

`=(s^2)/((s^2-9)^2)`

And we can easily find the remaining term's antiderivative by integrating by parts (for the last time!), or by simply exchanging
`\cosh` with
`\sinh` in the derivation of
`v_1`, so that we have


`\displaystyle\frac3{s^2-9}\int_0^\infty\sinh3te^(-st)\,\mathrm dt=\frac3{s^2-9}\left(-((s\sinh3t+3\cosh3t)e^(-st))/(s^2-9)\right)\bigg|_(t=0)^(t\to\infty)`

`=\frac9{(s^2-9)^2}`

(The exchanging is permissible because
`(\sinh x)'=\cosh x` and
`(\cosh x)'=\sinh x`; there are no alternating signs to account for.)

And so we conclude that


`\mathcal L_s\{t\cosh3t\}=(s^2+9)/((s^2-9)^2)`