TITLE: Finding the inverse Laplace transform of $\arctan \left(\frac{1}{s} \right)$using contour integration
QUESTION [10 upvotes]: Using contour integration, I want to show that $$\mathcal{L}^{-1} \left\{\text{arctan} \left(\frac{1}{s}\right) \right\}(x)= \frac{\sin x }{x}.$$
In other words, I want to show that $$ \frac{1}{2 \pi i} \int_{a - i \infty}^{a + i \infty} \text{arctan}\left(\frac{1}{s}\right) e^{xs} \ ds =\frac{\sin x}{x} ,$$ where $a$ is a constant greater than the real parts of all the singularities of $\text{arctan}\left(\frac{1}{s}\right)$.
We can define $\arctan \left(\frac{1}{s} \right)$ in terms of the complex logarithm.
Specifically, $$\arctan \left(\frac{1}{s} \right) = \frac{i}{2} \left[\log\left(1-\frac{i}{s}\right) - \log \left(1+\frac{i}{s}\right) \right]. $$
If we use the principal branch of the logarithm, then we need a branch cut on the imaginary axis from $-i$ to $i$.
I don't understand how to close contour.

REPLY [3 votes]: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
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\begin{align}
&\bbox[5px,#ffd]{\int_{0^{+}\ -\ \infty\ic}
^{0^{+}\ +\ \infty\ic}\arctan\pars{1 \over s}\expo{st}\,{\dd s \over 2\pi\ic}} =
\int_{0}^{1}\int_{0^{+}\ -\ \infty\ic}
^{0^{+}\ +\ \infty\ic}
{s\expo{st} \over s^{2} + x^{2}}\,{\dd s \over 2\pi\ic}\,\dd x
\\[5mm] & =
\int_{0}^{1}\pars{{-\ic x\expo{-\ic tx} \over -\ic x - \ic x} +
{\ic x\expo{-\ic tx} \over \ic x + \ic x}}\dd x =
\int_{0}^{1}\cos\pars{tx}\dd x
\\[5mm] = &\ \bbx{\ds{\sin\pars{t} \over t}} \\ &
\end{align}