Transforms of Integrals I have question regarding Laplacian transforms of integrals.  This set of problems does not allow for evaluation of the integral before transforming.  So I have a problem... $$\mathscr{L} \{t\int_0^t sinτdτ \}$$
The solution is $$\frac{3s^2+1}{s^{2}(s^{2}+1)^{2}}$$
The few worked solutions I have seen for this problem use a mish-mash of the theorems $\mathscr{L} \{t^{n}f(t)\}=(-1)^{n}\frac{d^{n}}{ds^{n}}F(s)$ and $\mathscr{L} \{\int_0^tf(τ)dτ \}=\frac{F(s)}{s}.$
My issue is with the logic behind this approach.  Perhaps I am looking at this too straightforwardly, but the first theorem above is written to suggest it holds only with functions of $t$.  The second, with $τ$.  If there is no direct way to convert a function of $τ$ to a function of $t$ within the LT operator, how are we able to superimpose a theorem combining elements of the two theorems to obtain the above solution?  Specifically, how are we able to achieve... $$\mathscr{L} \{t\int_0^t sinτdτ \}=-\frac{d}{ds}(\frac{1}{1+s^{2}})$$
... when $\mathscr{L} \{t\int_0^t sinτdτ \}$ is technically to be seen as $\mathscr{L} \{\int_0^t tsin(τ)dτ \}=\mathscr{L} \{\int_0^t f(t)g(τ)dτ \}$ with $f(t)=t$ and $g(τ)=sin(τ)$?  Again, without being allowed to evaulate the integral and with no way to convert $g(τ)$ to $g(t)$, it seems illogical to just use the theorems and hope they hold.  Am I overlooking something?  Please, help!
 A: Using 
$$ t \, e^{-st} = - \frac{d}{ds} \, e^{-st}$$
then 
\begin{align}
\int_{0}^{\infty} e^{-st} \, t \, f(t) \, dt &= - \frac{d}{ds} \, \int_{0}^{\infty} e^{- s t} \, f(t) \, dt.
\end{align}
Now considering the convolution theorem, in the form, where $\mathcal{L}\{f(t)\} \Doteq \overline{f}(s)$,
$$\overline{f}(s) \, \overline{g}(s) \Doteq \int_{0}^{t} f(t-u) \, g(u) \, du.$$
The Laplace transform of a constant is
$$\mathcal{L}\{a\} = \int_{0}^{\infty} e^{-st} \, a \, dt = \frac{a}{s}$$
or 
$$a \Doteq \frac{a}{s}.$$
Returning to the convolution set $f(t) = 1$ to obtain
$$\frac{\overline{g}(s)}{s} \Doteq \int_{0}^{t} g(u) \, du.$$
Since the parts have been established consider the transform:
\begin{align}
\mathcal{L}\left\{ t \int_{0}^{t} g(u) \, du \right\} &= \int_{0}^{\infty} e^{-st} \, t \, \int_{0}^{t} g(u) \, du \, dt \\
&= - \frac{d}{ds} \, \mathcal{L}\left\{ \int_{0}^{t} g(u) \, du \right\} \\
&= - \frac{d}{ds} \, \left( \frac{\overline{g}(s)}{s} \right) 
\end{align}
Two examples:
\begin{align}
\sin(a t) &\Doteq \frac{a}{s^{2} + a^{2}} \\
\cos(a t) &\Doteq \frac{s}{s^{2} + a^{2}}
\end{align}
\begin{align}
\mathcal{L}\left\{ t \int_{0}^{t} \sin(a u) \, du \right\} &= - \frac{d}{ds} \, \left( \frac{a}{s \, (s^{2} + a^{2})} \right) = \frac{a\, (3 s^{2} + a^{2})}{s^{2} \, (s^{2} + a^{2})^{2}} \\
\mathcal{L}\left\{ t \int_{0}^{t} \cos(a u) \, du \right\} &= - \frac{d}{ds} \, \left( \frac{1}{s^{2} + a^{2}} \right) = \frac{2 \, s}{(s^{2} + a^{2})^{2}}.
\end{align}
