Inverse Laplace transform of $\operatorname{arccot}(s)$, $\arctan(s)$ How would one find inverse Laplace transforms of  $\operatorname{arccot}(s)$ or of  $\arctan(s)$  without knowing in advance that this is related to  $\dfrac{\sin x}{x}$?
 A: The derivative of the arctangent is $\frac{d\arctan s}{ds}=\frac{1}{s^2+1}$.  Thus, we have
$$\begin{align}
\frac{1}{s^2+1}&=\frac{d}{ds}\int_0^{\infty}f(t)e^{-st}dt\\\\
&=-\int_0^{\infty}tf(t)e^{-st}dt
\end{align}$$
where the Laplace transform of $f$, $\mathscr{L}\left(f\right)(s)=\arctan(s)$ and the Laplace transform of $-tf$, $\mathscr{L}\left(-tf\right)(s)=\frac{1}{s^2+1}$.  We recall that the inverse Laplace transform of $\frac{1}{s^2+1}$ is indeed $\sin t$. Thus, we immediately see that $-tf(t)=\sin t$.
Now, we observe that for $t\ne 0$, $f(t)=-\frac{\sin t}{t}$.  We write $f$ as
$$f(t)=\mathscr{D}(t)-\frac{\sin t}{t}$$ 
where $\mathscr{D}(t)$ is a distribution that is $0$ for $t\ne 0$.
To find $\mathscr{D}(t)$, we simply note that the Laplace Transform of $-\frac{\sin t}{t}$ is $\arctan (s)-\frac{\pi}{2}$, from which we see immediately that $\mathscr{D}(t)=\frac{\pi}{2}\delta(t)$ and we are done.

An alternative way to find $\mathscr{D}(t)$ is a follows.  For any smooth test function $\phi(t)$, we have heuristically for "small" $\delta>0$
$$\begin{align}
\int_0^{\infty}f(t)\phi(t)dt&=\int_0^{\delta}\mathscr{D}(t)\phi(t)dt-\int_{0}^{\infty}\frac{\sin t}{t}\phi(t)dt\\\\
&\approx. \phi(0)\int_0^{\delta}\mathscr{D}(t)dt-\int_{0}^{\infty}\frac{\sin t}{t}\phi(t)dt\\\\
\end{align}$$
To find $\int_0^{\delta}\mathscr{D}(t)dt$ we need only observe that 
$$\begin{align}
\int_0^{\infty}f(t)dt&=\int_0^{\infty}\mathscr{D}(t)dt-\int_0^{\infty}\frac{\sin t}{t}dt\\\\
&=\int_0^{\infty}\mathscr{D}(t)dt-\frac{\pi}{2}\\\\
&=\arctan (0)\\\\
&=0 
\end{align}$$ 
Thus, $\int_0^{\infty}\mathscr{D}(t)dt=\int_0^{\delta}\mathscr{D}(t)dt =\frac{\pi}{2}$ for any $\delta >0$.  

Finally, we have
$$\bbox[5px,border:2px solid #C0A000]{f(t)=\frac{\pi}{2}\delta(t)-\frac{\sin t}{t}}$$
which is the inverse Laplace Transform of $\arctan (s)$!
