Trouble with Laplace Transform Can anyone help me with this Laplace Transform $$\mathcal{L}[(1-\cos(u))/u] ?$$
Thanks in advance
 A: Notice that
$$ \frac{1-\cos{u}}{u} = \int_0^1 \sin{u} \, da. $$
Interchanging the order of integration,
$$ \mathcal{L}\left( \frac{1-\cos{u}}{u} \right)(s) = \int_0^1 \int_0^{\infty} e^{-su} \cos{au} \, du \, da. $$
Now, we know the Laplace transform of $\cos{au}$, or at least we can get it using complex exponentials or integration by parts: it's $\frac{a}{a^2+s^2}$. Therefore the transform we need to evaluate is given by
$$ \int_0^1 \frac{a}{a^2+s^2} \, da. $$
But this is integral is easy: it's just
$$\left[ \frac{1}{2} \log{(a^2+s^2)} \right]_{a=0}^1 = \frac{1}{2} \log{\left( 1 + \frac{1}{s^2} \right)}. $$
A: Here is another way to proceed.  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 $(1-\cos(t))/t$ gives 
$$\begin{align}
\mathscr{L} \left( \frac{1- \cos t}{t}\right)(s)&=\int_0^{\infty} \left(\frac{1- \cos t}{t}\right)e^{-st}\,dt\\\\
&=\int_0^{\infty} (1- \cos t)\left(\frac{d\log(t)}{dt}\right)e^{-st}\,dt\\\\
&=\int_0^{\infty} \left(e^{-st}- \frac12 e^{-(s+i)t}-\frac12 e^{-(s-i)t}\right)\left(\frac{d\log(t)}{dt}\right)\,dt\\\\
&\overbrace{=}^{IBP}-\int_0^{\infty} \log(t)\frac{d}{dt}\left(e^{-st}- \frac12 e^{-(s+i)t}-\frac12 e^{-(s-i)t}\right) \,dt\\\\
&=s\int_0^{\infty} \log(t)e^{-st}\,dt\\\\
&-\frac12(s+i)\int_0^{\infty} \log(t)e^{-(s+i)t}\,dt\\\\
&-\frac12(s-i)\int_0^{\infty} \log(t)e^{-(s-i)t}dt\\\\
&\overbrace{=}^{\text{Using}\,(1)}s\left(\frac{-\gamma-\log(s)}{s}\right)\\\\
&-\frac12 (s+i)\left(\frac{-\gamma-\log(s+i)}{s+i}\right)\\\\
&-\frac12 (s-i)\left(\frac{-\gamma-\log(s-i)}{s-i}\right)\\\\
&=\frac12 \log\left(\frac{s^2+1}{s^2}\right)
\end{align}$$

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.
