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!
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}