# Finding the inverse Laplace transform of $\arctan \left(\frac{1}{s} \right)$using contour integration

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.

• Why do you want to show that by contour integration? It is much easier if you apply Laplace transform to $\frac{\sin t}{t}$. – Sungjin Kim Apr 1 '13 at 16:45
• Because we learn a lot about the structure of the function by working the inverse out directly. – Ron Gordon Apr 1 '13 at 17:06
• I see why contour integration is useful in obtaining inverse Laplace transform, but for this particular problem, using contour integration will be difficult. On the other hand, using Laplace transform directly to $\frac{\sin t}{t}$ is almost immediate. – Sungjin Kim Apr 2 '13 at 4:19
• That's precisely how you evaluate that integral; see this: math.stackexchange.com/questions/329347/… – Ron Gordon Apr 2 '13 at 20:15
• All I am saying is that you need to justify the signs you use by citing Cauchy's integral theorem. I didn't say you weren't clear or wrong. Just that, in forming the closed loop, you get the signs of the integrals relative to the original ILT. That's all. – Ron Gordon Apr 2 '13 at 20:39

I haven't worked out the integral along the contour yet, but here's a contour I had in mind:

This is definitely a sticky one.

• I was wondering how the cut could be surrounded but not enclosed inside of the contour. I often have a hard time visualizing stuff. So the integral along six parts of that contour go to zero in the limit, and you're left with the integral from $-i$ to $i$ just to the right of the cut and the integral from $i$ to $-i$ just to left of the cut? – Random Variable Apr 2 '13 at 16:19
• I take that back. The integrals above and below the negative real axis cancel each other since there is no cut along that axis. – Random Variable Apr 2 '13 at 16:27
• @RandomVariable: yes, you're right - no branch points along the neg. real axis, so the contributions cancel, I expect. The contributions should come from the vertical sections adjoining the branch cut. – Ron Gordon Apr 2 '13 at 16:32
• @RandomVariable: help me a little with that... – Ron Gordon Apr 2 '13 at 17:31
• I need to stew over this some. It may be a few days, but I will stew. – Ron Gordon Apr 2 '13 at 18:00

The following answer uses the contour provided above by Ron Gordon.

From the Maclaurin series of $\arctan(s)$, we can deduce that $\arctan \left(\frac{1}{s} \right) \sim \frac{1}{s}$ as $|s| \to \infty$.

We can then use a modified version of Jordan's lemma to conclude that $\int \arctan \left(\frac{1}{s} \right) e^{xs} \, ds$ vanishes along the two big arcs of the contour as the radii of the arcs go to infinity.

Also, since $\lim_{s \to \pm i} (s \mp i) \arctan \left( \frac{1}{s}\right)=0$, we get no contributions from letting the radii of the small circles about the branch points go to zero.

Moving clockwise around the branch point at $s=1$, the value of $\arctan \left( \frac{1}{s} \right)$ increases by $\pi$.

And moving clockwise around the branch point at $s=-i$, the value of $\arctan \left(\frac{1}{s} \right)$ decreases by $\pi$.

Therefore, \begin{align} \mathcal{L}^{-1} \left\{\arctan \left(\frac{1}{s} \right) \right\}(x) &= \frac{1}{2 \pi i} \int^{a+ i \infty}_{a - i \infty} \arctan \left(\frac{1}{s} \right) e^{xs} \, ds \\ &= - \frac{1}{2 \pi i} \left(\pi \int_{1}^{0} e^{ixt} \, i \, dt - \pi \int_{-1}^{0} e^{ixt} \, i \, dt \right) \\ &= \frac{1}{2} \int_{-1}^{1} e^{ixt} \, dt \\ &= \frac{1}{2ix} \left(e^{ix} - e^{-ix} \right) \\ &= \frac{\sin(x)}{x}. \end{align}

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \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 = \bbx{\ds{\sin\pars{t} \over t}} \end{align}