Laplace Question $f(t) = e^{-t} \sin(t)$

Question

I need help with this Laplace question. $$f(t) = e^{-t} \sin(t) $$

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Answer should be $\dfrac{1}{s^2 + 2s + 2}$

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What I'm currently doing is as follows:

$u = \sin(t)\qquad$ $dv = e^{-(s+1)t}dt$

$du = \cos(t)dt\qquad$ $v = \dfrac{e^{-(s+1)t}}{-(s+1)}$

$\dfrac{-\sin(t) e^{-(s+1)t}}{-(s+1)} - \int\dfrac{ e^{-(s+1)t}\cos(t)}{ -(s+1)} dt$

But even if I solved the integral, I wouldn't get this (which is what I should, see picture).

Answer 1

You need not integrate by parts to evaluate this integral. In fact, one would need to integrate by parts twice. See the section following the highlighted SPOILER ALERT

So, I thought it would be instructive to present a "trick" that we can use to quickly evaluate the integral of interest and other similar integrals.

HERE IS A HINT:

$$\sin(t)=\text{Im}\left(e^{it}\right) \tag 1$$

SPOILER ALERT: Scroll over the highlighted area to reveal the solution

Using $(1)$, we have$$\begin{align}\int_0^\infty e^{-t}\sin(t)e^{-st}\,dt&=\text{Im}\left(\int_0^\infty e^{(-1-s+i)t}\,dt\right)\\\\&=\text{Im}\left(\frac{1}{-(1+s)+i)}\right)\\\\&=\text{Im}\left(\frac{-(s+1)-i}{(s+1)^2+1}\,dt\right)\\\\&=\frac{1}{s^2+2s+2}\end{align}$$as was to be shown! Quick, painless, and less prone to careless errors.

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The last result in the OP was

$$\int_0^\infty e^{-t}\sin(t)e^{-st}\,dt=\left.\left(\frac{-\sin(t) e^{-(s+1)t}}{-(s+1)}\right)\right|_0^\infty - \int_0^{\infty}\dfrac{ e^{-(s+1)t}\cos(t)}{ -(s+1)} dt$$

Continuing we have

$$\begin{align} \int_0^\infty e^{-t}\sin(t)e^{-st}\,dt&=\frac{1}{s+1} \int_0^{\infty} e^{-(s+1)t}\cos(t)\,dt \tag 2 \end{align}$$

We integrate by parts $(2)$ with $u=\cos(t)$ and $v=-\frac{e^{-(s+1)t}}{s+1}$. Then, we obtain

$$\begin{align} \int_0^\infty e^{-t}\sin(t)e^{-st}\,dt&=\frac{1}{s+1}\left.\left(-\frac{\cos(t)e^{-(s+1)t}}{s+1}\right)\right|_0^\infty -\frac{1}{(s+1)^2}\int_0^\infty \sin(t)e^{-(s+1)t}\,dt\\\\ &=\frac{1}{(s+1)^2}-\frac{1}{(s+1)^2}\int_0^\infty \sin(t)e^{-(s+1)t}\,dt\\\\ ((s+1)^2+1)\int_0^\infty e^{-t}\sin(t)e^{-st}\,dt&=1\\\\ \int_0^\infty e^{-t}\sin(t)e^{-st}\,dt&=\frac{1}{s^2+2s+2} \end{align}$$

as expected!

Answer 2

Calculate Laplace transform for $f(t)$ and $e^{-t}\cos(t)$ then you'll get a equation system where the unknowns will be the integrals $$\int\dfrac{ e^{-(s+1)t}\cos(t)}{ -(s+1)} dt \text{ and } \int\dfrac{ e^{-(s+1)t}\sin(t)}{ -(s+1)} dt $$ with this you can conclude.

Answer 3

Here's the 'direct' proof: $f(t)=e^{-t}\sin t.$ \begin{eqnarray*}\mathcal{L}\{f(t)\}&=&\int_{0}^{+\infty}e^{-(s+1)t}\sin t\,dt~=~\lim_{\ell\to+\infty}\int_{0}^{\ell}e^{-(s+1)t}\sin t\,dt\\ &=&-\frac{1}{s+1}\lim_{\ell\to+\infty}\left[e^{-(s+1)t}\sin t\Big|_{0}^{\ell}-\int_{0}^{\ell}e^{-(s+1)t}\cos t\,dt\right]\\ &=&\frac{1}{s+1}\lim_{\ell\to+\infty}\int_{0}^{\ell}e^{-(s+1)t}\cos t\,dt, \end{eqnarray*}
since $$\lim_{\ell\to+\infty}e^{-(s+1)\ell}\sin\ell=0.$$ Integrating again by parts yields \begin{eqnarray*}\mathcal{L}\{f(t)\}&=&-\frac{1}{(s+1)^2}\lim_{\ell\to+\infty}\left[e^{-(s+1)t}\cos t\Big|_{0}^{\ell}+\int_{0}^{\ell}e^{-(s+1)t}\sin t\,dt\right]\\ &=&\frac{1}{(s+1)^2}\left[1-\lim_{\ell\to+\infty}\int_{0}^{\ell}e^{-(s+1)t}\sin t\,dt\right]\\ &=&\frac{1}{(s+1)^2}(1-\mathcal{L}\{f(t)\}) \end{eqnarray*}
and finally solving by $\mathcal{L}\{f(t)\}$, $$\boxed{\mathcal{L}\{f(t)\}=\frac{1}{(s+1)^2+1}=\frac{1}{s^2+2s+2}}~.$$