Finding the inverse Laplace transform of $ \ln \! \left( 1 + \frac{1}{s^{2}} \right) $. Can someone help me find the inverse Laplace transform of $ \ln \! \left( 1 + \dfrac{1}{s^{2}} \right) $? I have no idea where to start.
 A: 
The idea is to undo the operations you may find in the transform using the properties of the Laplace transform. The Laplace transform of the logarithm times Heaviside is essentially a logarithm divided by $s$. The transform of a derivative is essentially the transform of the function times $s$. And finally the transform of a function multiplied by an exponential is the transform of the function translated.


We can write 
$$\begin{align}\ln\left(1+\frac{1}{s^2}\right)&=\ln(s+i)+\ln(s-i)-2\ln(s)\\&=(s+i)\frac{\ln(s+i)+\gamma}{s+i}+(s-i)\frac{\ln(s-i)+\gamma}{s-i}-2s\frac{\ln(s)+\gamma}{s}\end{align}$$
We know that $$\mathcal{L}(-\ln(t)u(t))=\frac{\ln(s)+\gamma}{s}$$
Therefore 
$$\mathcal{L}\left(D\left(-\ln(t)u(t)\right)\right)=s\frac{\ln(s)+\gamma}{s}$$
where $D$ is derivative and 
$$\mathcal{L}\left(e^{\pm it}D\left(-\ln(t)u(t)\right)\right)=(s\pm i)\frac{\ln(s\pm i)+\gamma}{s\pm i}$$
Now you can complete it.
A: 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 
$$\begin{align}
\Gamma(z) &= \int_0^{\infty} (st)^{z-1}e^{-st}sdt\\\\
&=s^z\int_0^{\infty} t^{z-1}e^{-st}dt
\end{align}$$
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 $\log (s)$ yields
$$\log(s) = -\gamma -s\,\mathscr{L}\{\log (t)\}(s)$$
where we have noted that $\Gamma'(1)=-\gamma$, is the Euler-Mascheroni constant.
Now, following the decomposition given by Alamos, we have that
$$\begin{align}
\log\left(\frac{s^2+1}{s^2}\right)&=\log(s+i)+\log(s-i)-2\log(s)\\\\
&=-(s+i)\int_0^{\infty} \log(t)e^{-(s+i)t}dt-(s-i)\int_0^{\infty} \log(t)e^{-(s-i)t}dt+2s\int_0^{\infty} \log(t)e^{-st}dt\\\\
&=\int_0^{\infty} \log(t)\frac{d}{dt}(e^{-(s+i)t})dt+\int_0^{\infty} \log(t)\frac{d}{dt}(e^{-(s-i)t})dt-2\int_0^{\infty} \log(t)\frac{d}{dt}(e^{-st})dt\\\\
&=2\int_0^{\infty}\log(t)\frac{d}{dt}((\cos(t)-1)e^{-st})dt\\\\
&=2\int_0^{\infty}\frac{1-\cos(t)}{t}e^{-st}dt\\\\
&=2\mathscr{L}\left(\frac{1-\cos(t)}{t}\right)(s)
\end{align}$$
Inverting both sides reveals
$$\mathscr{L}^{-1}\left(\log\left(\frac{s^2+1}{s^2}\right)\right)(t)=2\,\,\frac{1-\cos(t)}{t}$$
A: Since $\mathscr{L}^{-1}\left\{ s^{-n} \right\}=\frac{1}{(n-1)!} t^{n-1}$, we have, in wiev of the expansion $\ln(1-x)=-\sum_{n=1}^\infty \frac{x^n}{n}$:
$$\mathscr{L}^{-1}\left\{\ln\left(1+\frac{1}{s^2}\right) \right\}=-\sum_{n=1}^{\infty}\frac{(-1)^n}{n}\mathscr{L}^{-1}\left\{s^{-2n} \right\}=-\sum_{n=1}^{\infty}\frac{(-1)^n}{n}\frac{t^{2n-1}}{(2n-1)!}.$$
The last sum can be arranged:
$$-\sum_{n=1}^{\infty}\frac{(-1)^n}{n}\frac{t^{2n-1}}{(2n-1)!}=-2\sum_{n=1}^{\infty}(-1)^n\frac{t^{2n-1}}{(2n)!}=\frac{2}{t}-\frac{2}{t}\sum_{n=0}^{\infty}(-1)^n\frac{t^{2n}}{(2n)!}=\frac{2}{t}(1-\cos t).$$
Thus proving the formula.
A: Use the following :
$$x(t)\to X(s)$$
$$-tx(t) \to X'(s)$$
Now we write
$$X(s) = \ln (1+1/s^2) = \ln(s^2+1) - 2\ln s$$
$$X'(s)  = \frac{2s}{s^2+1}-\frac{2}{s}$$
$$-tx(t) = 2\cos t -2\to x(t) = \frac{2-cos t}{t}$$
