Laplace Transform to evaluate $\int_{0}^{\infty} e^{-2t} *t*\sin(4t)dt$ The question is, how can I find the value of the integral $$\int_{0}^{\infty} e^{-2t} *t*\sin(4t)dt$$
I thought I could solve it by saying this is $L(t\sin(4t))(2)$. Since $L(t\sin(4t)) = \frac{8s}{(s^2+16)^2}$ we have that the integral is this at $s = 2$, or that it is $\frac{1}{25}$. I know this is wrong, but why?
EDIT: just checked wolfram alpha, it's right. But is it mathematically rigorous? (My instructor gave a MUCH longer answer).
 A: It's rigorous (IMO), and also very clever!
See, instead of using integration by parts, if you know about the Lapalce transform, you can use it! Just because a person's solution spans multiple pages does not mean another approach is "wrong." I always tell my students that sometimes going the long way means you're doing it right (in some contexts).
In general, if $f(t)$ admits a Laplace transform, then you can avoid integration by parts in integrating $f$ against $e^{-st}$ via the Laplace transform.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
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OP question: "The question is, how can I find the value of the integral ?":

\begin{align}
\int_{0}^{\infty}\expo{-2t}t\sin\pars{4t}\,\dd t & =
\left.-\,\partiald{}{\mu}\int_{0}^{\infty}\expo{-2t}\cos\pars{\mu t}\,\dd t\,
\right\vert_{\ \mu\ =\ 4} =
\left.-\,\partiald{}{\mu}\Re\int_{0}^{\infty}\expo{-\pars{2 - \ic\mu}t}\,\dd t\,
\right\vert_{\ \mu\ =\ 4}
\\[5mm] & =
\left.-\,\partiald{}{\mu}\Re{1 \over 2 - \ic\mu}\,\right\vert_{\ \mu\ =\ 4} =
\left.-\,\Re{\ic \over \pars{2 - \ic\mu}^{2}}\,\right\vert_{\ \mu\ =\ 4} =
\left.\Im{\pars{2 + \ic\mu}^{2} \over \pars{4 + \mu^{2}}^{2}}
\,\right\vert_{\ \mu\ =\ 4}
\\[5mm] & =
\left.{4\mu \over \pars{4 + \mu^{2}}^{2}}
\,\right\vert_{\ \mu\ =\ 4} = {4 \times 4 \over \pars{4 + 4^{2}}^{2}} =
\bbx{1 \over 25}
\end{align}
A: Well, in general:
$$\mathscr{L}_t\left[t\sin\left(\text{n}t\right)\right]_{\left(\text{s}\right)}:=\int_0^\infty t\sin\left(\text{n}t\right)\cdot e^{-\text{s}t}\space\text{d}t\tag1$$
We can use the 'frequency-domain derivative' property of the Laplace transform:
$$\mathscr{L}_t\left[t\sin\left(\text{n}t\right)\right]_{\left(\text{s}\right)}=-\frac{\partial}{\partial\space\text{s}}\space\left(\mathscr{L}_t\left[\sin\left(\text{n}t\right)\right]_{\left(\text{s}\right)}\right)=-\frac{\partial}{\partial\space\text{s}}\space\left\{\int_0^\infty\sin\left(\text{n}t\right)\cdot e^{-\text{s}t}\space\text{d}t\right\}\tag2$$
Now, the Laplace transform of the sine function can be found here:
$$\mathscr{L}_t\left[t\sin\left(\text{n}t\right)\right]_{\left(\text{s}\right)}=-\frac{\partial}{\partial\space\text{s}}\space\left\{\frac{\text{n}}{\text{s}^2+\text{n}^2}\right\}=\frac{2\text{n}\cdot\text{s}}{\left(\text{s}^2+\text{n}^2\right)^2}\tag3$$
