Laplace Tansform of $\frac{1-\cos(2t)}t$ What is the Laplace Transform of $\frac{1-\cos(2t)}t$ and is there a general formula for such situations with $\cos$, $\sin$ and others?
 A: Note that
$$\begin{align}\frac{1-\cos\omega t}{t}&=\int_0^{\omega}\sin\omega't\,\mathrm{d}\omega' \\
\frac{\sin\omega t}{t}&=\int_0^{\omega}\cos\omega't\,\mathrm{d}\omega'
\end{align}$$
so that, once you've justified switching the order of integration, you can get these functions' Laplace transforms by integrating those for $\cos\omega t$ and $\sin \omega t$ with respect to $\omega$.
A: We will use the identity 
$$\mathscr{L}\{\log (t)\}(s)=\frac{-\gamma-\log(s)}{s}\tag1$$
where $\mathscr{L}\{\log (t)\}(s)$ is the Laplace Transform of $\log (t)$ and where $\gamma$ is the Euler-Mascheroni constant.  We will prove this identity at the end of this writing.
Now, taking the Laplace Transform of $(1-\cos(2t))/t$ gives 
$$\begin{align}
\mathscr{L} \left( \frac{1- \cos (2t)}{t}\right)(s)&=\int_0^{\infty} \left(\frac{1- \cos (2t)}{t}\right)e^{-st}\,dt\\\\
&\overbrace{=}^{t\mapsto t/2}\int_0^{\infty} \left(\frac{1- \cos (t)}{t}\right)e^{-st/2}\,dt\\\\
&=\int_0^{\infty} (1- \cos (t))\left(\frac{d\log(t)}{dt}\right)e^{-st/2}\,dt\\\\
&=\int_0^{\infty} \left(e^{-st/2}- \frac12 e^{-(s/2+i)t}-\frac12 e^{-(s/2-i)t}\right)\left(\frac{d\log(t)}{dt}\right)\,dt\\\\
&\overbrace{=}^{IBP}-\int_0^{\infty} \log(t)\frac{d}{dt}\left(e^{-st/2}- \frac12 e^{-(s/2+i)t}-\frac12 e^{-(s/2-i)t}\right) \,dt\\\\
&=\frac s2\int_0^{\infty} \log(t)e^{-st/2}\,dt\\\\
&-\frac12(s/2+i)\int_0^{\infty} \log(t)e^{-(s/2+i)t}\,dt\\\\
&-\frac12(s/2-i)\int_0^{\infty} \log(t)e^{-(s/2-i)t}dt\\\\
&\overbrace{=}^{\text{Using}\,(1)}s\left(\frac{-\gamma-\log(s)}{s}\right)\\\\
&-\frac12 (s/2+i)\left(\frac{-\gamma-\log(s/2+i)}{s/2+i}\right)\\\\
&-\frac12 (s/2-i)\left(\frac{-\gamma-\log(s/2-i)}{s/2-i}\right)\\\\
&=\frac12 \log\left(\frac{s^2+4}{s^2}\right)
\end{align}$$

Proof of the identity
$$\mathscr{L}\{\log (t)\}(s)=-\frac{\gamma-\log(s)}{s}$$
The Gamma Function $\Gamma$ is defined as 
$$\Gamma(z) = \int_0^{\infty} t^{z-1}e^{-t}\,dt$$
for $\text{Re}\{z\}>0$.
We can write this integral representation as a Laplace Transform by letting $t \to st$.  Then, we have 
$$\begin{align}
\Gamma(z) &= \int_0^{\infty} (st)^{z-1}e^{-st}s\,dt\\\\
&=s^z\int_0^{\infty} t^{z-1}e^{-st}\,dt
\end{align}$$
The derivative of the Gamma Function follows directly as
$$\begin{align}
\Gamma'(z) &= s^z\log (s)\int_0^{\infty} t^{z-1}e^{-st}dt+s^z\int_0^{\infty} t^{z-1}e^{-st}\log (t) \,dt
\end{align}$$
Note that $\Gamma'(1)=\log(s)+s\mathscr{L}\{\log (t)\}(s)$, where 
$$\mathscr{L}\{\log (t)\}(s)=\int_0^{\infty}\log(t) e^{-st}\,dt$$
is the Laplace Transform of $\log (t)$. Solving for $\mathscr{L}\{\log (t)\}(s)$ yields
$$\mathscr{L}\{\log (t)\}(s)= \frac{-\gamma-\log(s)}{s}$$
where we have noted that $\Gamma'(1)=-\gamma$, is the Euler-Mascheroni constant.
