Inverse Laplace transform of $\tan^{−1}\left(\frac{1}{s}\right)$

Question

I'm studying Laplace transformations, but I don't understand where $-\frac{1}{t}$ comes from. And what is the relationship between the corollary and the example?

Answer 1

Let $F(s)=\mathscr{L}\{f(t)\}(s)$. Then,

$$\begin{align} \mathscr{L}\{tf(t)\}&=\int_0^\infty f(t)\left(te^{-st}\right)\,dt\\\\ &=\int_0^\infty f(t)\left(-\frac{de^{-st}}{ds}\right)\,dt\\\\ &=-\frac{d}{ds}\left(\int_0^\infty f(t)e^{-st}\,dt\right)\\\\ &=-\frac{d}{ds}\left(F(s)\right) \tag 1 \end{align}$$

We can continue inductively to prove the "Corollary" in the OP.

For the case at hand, we have $F(s)=\arctan(1/s)$. Therefore,

$$-\frac{dF(s)}{ds}=\frac{1}{s^2+1}$$

Recognizing that $\frac{1}{s^2+1}$ is the Laplace Transform of $\sin(t)$, and using the result from $(1)$ gives

$$\mathscr{L}\{tf(t)\}=\mathscr{L}\{\sin(t)\}$$

from which we see immediately that $f(t)=\frac{\sin(t)}{t}$. And we are done!

Answer 2

Think of $\tan^{-1} \left( \frac 1 s\right)$ as the antiderivative of $\frac{-1}{s^2+1}$.

Then in your example, $n=1$. This introduces a factor of $1/t$ on the left hand side, and a negative sign on the right hand side.