Inverse Laplace transform $\mathcal{L}^{-1}\left \{ \ln \left ( 1+\frac{w^{2}}{s^{2}}\right ) \right \}$ Where $s\in \mathbb{C}$. I assume that this would be pretty easily handled by solving it by definition, but I haven't taken courses in complex analysis yet. Also, I can't think of any nice property of the Laplace transform that would yield the answer. At second glance, a series represnation might do the trick, but then I remembered that s is a complex variable and thus the function can only be represented via (complex) Laurent series.
EDIT:
I am looking for an approach different from what copper.hat suggested in the comments, namely using some of the nice properties of the transform.
 A: Let's write down the expression for the inverse transform:
$$\frac1{i 2 \pi} \int_{c-i \infty}^{c+i \infty} ds \, \log{\left (1+\frac{w^2}{s^2} \right )} e^{s t} $$
where $c$ is greater than the largest real part of all the singularities of the integrand, if any.  
To compute the inverse transform above, consider the contour integral for $t \gt 0$:
$$\oint_{\gamma} dz \, \log{\left (1+\frac{w^2}{z^2} \right )} e^{z t} $$
where $\gamma$ is the contour illustrated below.

Note that we have deformed the contour so as to avoid the branch points at $z=\pm i w$. The outer circular arc has radius $R$ and the arcs about the branch points have radius $\epsilon$.
I will for the time being assure the reader that the integrals over the outer arcs vanish in the limit as $R \to \infty$, and the integrals over the inner arcs vanish in the limit as $\epsilon \to 0$.  Therefore, we consider the contour integral as the limit as $R \to \infty$ and $\epsilon \to 0$.
To do this, we must look carefully at the behavior of the log term on the dog-bone portion of the contour.  Note that
$$\log{\left (1+\frac{w^2}{z^2} \right )} = \log{\left (1+i \frac{w}{z} \right )}+\log{\left (1-i\frac{w}{z} \right )} $$
We take the log to be the principal branch, so that the branch cut occurs for the arg of the arguments of the log equal to $\pi$. 
The reader should note that there is a fourfold split here: when $z$ is to the left or right of the imaginary axis, and when $z$ is above or below the real axis.  In each of these cases, the log has a negative argument, but the log of the negative argument takes on a different value along each section of the dog-bone.
When $z$ is to the left and right of the imaginary axis we respectively parametrize $z=+i y$and $z=-i y$.  However, we subtract $i 2 \pi$ from the log term upon crossing the imaginary axis and add $i 2 \pi$ upon crossing the real axis.
The contour integral is then equal to 
$$\int_{c-i \infty}^{c+i \infty} ds \, \log{\left (1+\frac{w^2}{s^2} \right )} e^{s t} + i \int_0^w dy \, \left [\log{\left (\frac{w^2}{y^2}-1 \right )}+i \pi \right ] e^{i y t}\\ -  i \int_w^0 dy \, \left [\log{\left (\frac{w^2}{y^2}-1 \right )}-i \pi \right ] e^{-i y t} -  i \int_0^w dy \, \left [\log{\left (\frac{w^2}{y^2}-1 \right )}+i \pi \right ] e^{-i y t}\\+ i \int_w^0 dy \, \left [\log{\left (\frac{w^2}{y^2}-1 \right )}-i \pi \right ] e^{i y t}$$
Note that the log terms all cancel, and we are left with
$$\int_{c-i \infty}^{c+i \infty} ds \, \log{\left (1+\frac{w^2}{s^2} \right )} e^{s t} + i 2 \pi (i) (i 2) \int_0^w dy \, \sin{y t} $$
By Cauchy's theorem, the contour integral is zero.  Therefore, we may immediately write down the inverse transform as
$$\frac1{i 2 \pi} \int_{c-i \infty}^{c+i \infty} ds \, \log{\left (1+\frac{w^2}{s^2} \right )} e^{s t} = 2 \frac{1-\cos{w t}}{t} $$
A: $$\ln\left( 1+\frac{w^2}{s^2}\right)=-\displaystyle\sum_{k=1}^\infty \frac{(-1)^k w^{2k}}{k s^{2k}} $$
The inverse Laplace transform of $\frac{1}{s^{2k}}$ is $\frac{t^{2k-1}}{(2k-1)!}$
So, the inverse laplace transform of $\displaystyle\sum_{k=1}^\infty \frac{(-1)^k w^{2k}}{k s^{2k}} $ is $-\displaystyle\sum_{k=1}^\infty \frac{(-1)^k w^{2k}t^{2k-1}}{k (2k-1)!} $ which is equal to :
$$-\frac{1}{t}\displaystyle\sum_{k=1}^\infty \frac{(-1)^k w^{2k}t^{2k}(2k)}{k (2k)(2k-1)!}=-\frac{2}{t}\displaystyle\sum_{k=1}^\infty \frac{(-1)^k w^{2k}t^{2k}}{ (2k)!}=-\frac{2}{t}\left(-1+\displaystyle\sum_{k=0}^\infty \frac{(-1)^k w^{2k}t^{2k}}{ (2k)!}\right)$$
Since $\displaystyle\sum_{k=0}^\infty \frac{(-1)^k (wt)^{2k}}{ (2k)!}=\cos(wt)$ ,
finally, the Laplace inverse is :
$$\frac{2}{t}\left(1-\cos(wt)\right)$$
