Inverse Laplace transform of $\operatorname{arccot}(s)$, $\arctan(s)$

Question

How would one find inverse Laplace transforms of $\operatorname{arccot}(s)$ or of $\arctan(s)$ without knowing in advance that this is related to $\dfrac{\sin x}{x}$?

Answer 1

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.

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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 $\delta >0$.

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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 $\arctan (s)$!