How to do inverse Laplace transformation to $\arctan(s)$

Question

First I tried to make $ \arctan(s) = \frac{\pi}{2}-\int_s^{+\infty} \frac{1}{s^2+1}$ and then I can transform

$$ \frac{\pi}{2}\rightarrow \frac{\pi}{2}\delta(t), \qquad\int_s^{+\infty} \frac{1}{s^2+1}\rightarrow \frac{\sin(t)}{t}$$ And I get the answer $\frac{\pi}{2}\delta(t)-\frac{\sin(t)}{t}$. But the answer in my paper should be $-\frac{\sin(t)}{t}$ and I don't know why.

Answer 1

The inverse Laplace Transform of $F(s)=\arctan(s)$ as given by the Bromwich integral

$$\begin{align} \mathscr{L}^{-1}\{\arctan(s)\}(t)&=\frac{1}{2\pi i}\int_{\sigma-i\infty}^{\sigma+i\infty}\arctan(s)e^{st}\,ds\\\\ &=\frac{1}{4\pi}\int_{\sigma-i\infty}^{\sigma+i\infty}\left(\log(i+s)-\log(i-s) \right)e^{st}\,ds\,\tag 1 \end{align}$$

where $\sigma>0$, does not exist (Note that $\lim_{s\to \pm \infty}\arctan(s)= \pm \pi/2$).

We can evaluate, instead, the inverse Laplace Transform of $\arctan(s) - \pi/2$. Note that

$$\begin{align} \mathscr{L}^{-1}\{\arctan(s)-\pi/2\}(t)&=\frac{1}{2\pi i}\int_{\sigma-i\infty}^{\sigma+i\infty}(\arctan(s)-\pi/2)e^{st}\,ds\\\\ &=\frac{1}{4\pi}\int_{\sigma-i\infty}^{\sigma+i\infty}\left(\log(s+i)-\log(s-i) \right)e^{st}\,ds\tag 1 \end{align}$$

for $\sigma>0$.

To do so, we cut the plane with two branch cuts, which emanate from the branch points at $s=\pm i$ and extends to $-\infty \pm i$ along rays parallel to the real axis. Then for $t>0$, applying Cauchy's Integral Theorem to the integral in $(1)$ reveals

$$\begin{align} \int_{\sigma-i\infty}^{\sigma+i\infty}(\arctan(s)-\pi/2)e^{st}\,ds&=\lim_{\epsilon\to 0^+}\left(\int_0^{-\infty}\left(\log(x+i2)-\log(x+i\epsilon) \right)e^{xt+it}\,dx\\\\+\int_{-\infty}^0\left(\log(x+i2)-\log(x-i\epsilon) \right)e^{xt+it}\,dx\\\\ +\int_0^{-\infty}\left(\log(x+i\epsilon)-\log(x-i2) \right)e^{xt-it}\,dx\\\\ +\int_{-\infty}^0\left(\log(x-i\epsilon)-\log(x-i2) \right)e^{xt-it}\,dx\right)\\\\ &=-i2\pi\, e^{it}\,\int_0^{-\infty}e^{xt}\,dx+i2\pi\,e^{-it}\,\int_0^{-\infty}e^{xt}\,dx\\\\ &=2\pi i \frac{e^{it}-e^{-it}}{t}\\\\ &=-4\pi \frac{\sin(t)}{t} \tag 2 \end{align}$$

whence the inverse Laplace Transform for $\arctan(s)-\pi/2$ is for $t>0$

$$\mathscr{L}^{-1}\{\arctan(s)-\pi/2\}(t)=-\frac{\sin(t)}{t}$$

Answer 2

The answer in the paper is wrong. The function

$$f \colon t \mapsto \begin{cases}\quad -1 &, t = 0 \\ -\dfrac{\sin t}{t} &, t \neq 0\end{cases}$$

is continuous and bounded (on $\mathbb{R}$), hence its Laplace transform $F$ (which is defined on the half-plane $\operatorname{Re} s > 0$) satisfies

$$\lim_{\operatorname{Re} s \to +\infty} F(s) = 0\tag{$\ast$}$$

by the dominated convergence theorem.

No branch of $\arctan$ satisfies that condition.

By differentiating under the integral, we find that

$$F'(s) = \int_0^{\infty} \sin t e^{-st}\,dt = \frac{1}{1+s^2},$$

from which it follows that

$$F(s) = C + \arctan s$$

for some constant $C$ and some branch of $\arctan$. Taking the principal branch of $\arctan$, by condition $(\ast)$ we find that we need to take $C = -\frac{\pi}{2}$. It seems the paper erroneously assumed that the choice $C = 0$ was legitimate.