# Silly question about inverse Laplace transform

I know that the Laplace transform of $e^{at}$ is $\frac{1}{s-a}$ for $s>a$. This is easy.

This implies that the inverse Laplace transform of $\frac{1}{s-a}$ is $e^{at}$. Fine.

However, the theory says that the inverse Laplace transform of $F(s)$ is given by $$\mathcal{L}^{-1}(F(s))=\frac{1}{2\pi i}\int_{\gamma-i\infty}^{\gamma+i\infty} e^{st}F(s)ds,$$ where $\gamma$ is greater that the real part of all singularities of $F(s)$.

To apply this theory to $\frac{1}{s-a}$ I need $\gamma>a$. Now my question is: how do I close the contour in the complex plane to evaluate this integral?

(My first instinct is always to try a semi-circle, but I don't see how that would work in this case)

• You don't close the contour. The integral is over the line $\Re z=\gamma$. – Kavi Rama Murthy Jun 16 '18 at 22:58
• I thought the idea would be to start with a closed contour, so that the integral would be equal to the residue, and then take a limit when one arc of the contour approached that line. Is this the wrong idea, then? – thedude Jun 16 '18 at 23:19
• The point of the trick with the contour is that you can use residue theory. Your integral is then "Integral over line + Integral over arc = Integral over contour". So, if you can compute the integral over the contour by residue theory, and the integral over the arc (or other shape) goes to 0 as the arc gets bigger, then the integral over the contour is the integral over the line. You usually choose the arc such that it never crosses singularities as you take the limit to infinity. – Raskolnikov Jun 17 '18 at 5:44
• @Raskolnikov Yes, I know that much, but what contour would you choose for this particular problem? – thedude Jun 17 '18 at 11:51
• You can use a large semicircle to the left of $\operatorname{Re} s = \gamma$. By Jordan's lemma the integral over that tends to $0$ if the radius tends to $+\infty$. Or you can take a rectangular contour with vertices $\gamma \pm iT$ and $-S \pm iT$, letting $S, T \to +\infty$. The integral over the part closing the rectangular contour is easy to bound using $\lvert e^{st}\rvert = e^{t\operatorname{Re} s}$. – Daniel Fischer Jun 17 '18 at 14:57

## 1 Answer

We can close the contour by a circular arc to the left of the line $\operatorname{Re} s = \gamma$, or we can use straight lines to make a rectangular contour. Let's first look at the rectangular method. Let $S > \lvert a\rvert$ (so that the contour is sure to enclose the singularity at $a$) and $T > 0$. Then consider the rectangular contour $R_{S,T}$ with vertices $\gamma - iT$, $\gamma + iT$, $-S + iT$, $-S - iT$. By the residue theorem we have

$$\frac{1}{2\pi i} \int_{R_{S,T}} \frac{e^{st}}{s - a}\,ds = \operatorname{Res}\biggl(\frac{e^{st}}{s-a}; a\biggr) = e^{at}\,.$$

We estimate the contribution from the auxiliary line segments and see that this tends to $0$ as $S \to +\infty$ and $T \to +\infty$. On the segment from $-S + iT$ to $-S - iT$ we have $\lvert s-a\rvert \geqslant S - \lvert a\rvert$ and $\lvert e^{st}\rvert = e^{t\operatorname{Re} s} = e^{-St}$. Thus

$$\Biggl\lvert \int_{-S+iT}^{-S-iT} \frac{e^{st}}{s-a}\,ds\Biggr\rvert \leqslant 2T\cdot \frac{e^{-St}}{S - \lvert a\rvert} \leqslant \frac{2T}{S - \lvert a\rvert}$$

by the standard estimate.

On the horizontal segments from $\gamma + iT$ to $-S + iT$ and from $-S - iT$ to $\gamma - iT$ we have $\lvert s-a\rvert \geqslant \lvert \operatorname{Im} (s-a)\rvert = T$ (this is for $a$ real, for complex $a$ we can work with $T - \lvert a\rvert$) and we can hence estimate

\begin{align} \Biggl\lvert \int_{-S-iT}^{\gamma-iT} \frac{e^{st}}{s-a}\,ds\Biggr\rvert &\leqslant \frac{1}{T}\int_{-S}^{\gamma} e^{\sigma t}\,d\sigma \\ &< \frac{1}{T} \int_{-\infty}^{\gamma} e^{\sigma t}\,d\sigma \\ &= \frac{e^{\gamma t}}{Tt} \end{align}

for $t > 0$. We have the same bound for the horizontal segment from $\gamma + iT$ to $-S + iT$. Together, these estimates yield

$$\Biggl\lvert \frac{1}{2\pi i}\int_{\gamma - iT}^{\gamma + iT} \frac{e^{st}}{s-a}\,ds - e^{at}\Biggr\rvert \leqslant \frac{T}{\pi(S - \lvert a\rvert)} + \frac{e^{\gamma t}}{\pi Tt}$$

for all $t > 0$, $T > 0$ and $S > \lvert a\rvert$. Picking for example $S = T^2 + \lvert a\rvert$ the bound tends to $0$ for $T \to +\infty$, and we find

$$\frac{1}{2\pi i} \int_{\gamma - i\infty}^{\gamma + i\infty} \frac{e^{st}}{s-a}\,ds = e^{at} \tag{\ast}$$

for all $t > 0$. (In the argument, I have used a contour symmetric about the real axis, so the existence of the integral is only ascertained as a principal value integral. Showing that the integral exists as an improper Riemann integral is not more difficult, one considers a contour with vertices $\gamma - iU$, $\gamma + iT$, $-S + iT$, $-S - iU$. Then one first lets $S \to +\infty$, and afterwards independently $T \to +\infty$, $U \to +\infty$. The integral does not exist as a Lebesgue integral for $F(s) = (s-a)^{-1}$.)

If we close the contour by a circular arc, it is more convenient to have the centre of the circle at $0$ than to have it at $\gamma$, but the latter also works. Only the estimates are a bit more cumbersome. We split the arc into the part in the left half-plane and the part in the right half-plane (the latter is empty if $\gamma \leqslant 0$). If $\gamma > 0$, we estimate the part in the right half-plane by the standard estimate. We have $\lvert e^{st}\rvert \leqslant e^{\gamma t}$ and $\lvert s - a\rvert \geqslant \lvert \operatorname{Im} s\rvert$. Taking the radius $R \geqslant 2\gamma$ one finds that $\lvert \operatorname{Im} s \rvert \geqslant \frac{\sqrt{3}}{2}R$ on the part of the arc in the right half-plane, and the length of each of the two sub-arcs there is less than $\pi\gamma$, so

$$\Biggl\lvert \int\limits_{\substack{\lvert s\rvert = R \\ 0 < \operatorname{Re} s \leqslant \gamma}} \frac{e^{st}}{s-a}\,ds\Biggr\rvert \leqslant \frac{4\pi e^{\gamma t}}{\sqrt{3}\,R}\,.$$

For the part in the left half-plane, we use $\lvert s-a\rvert \geqslant R - \lvert a\rvert$ to estimate

$$\Biggl\lvert\int\limits_{\substack{\lvert s\rvert = R \\ \operatorname{Re} s \leqslant 0}} \frac{e^{st}}{s-a}\,ds\Biggr\rvert \leqslant \frac{R}{R - \lvert a\rvert} \int_0^{\pi} e^{-Rt\sin \varphi}\,d\varphi$$

using the parametrisation $s = iRe^{i\varphi}$. By the symmetry of the sine, and $\sin \varphi \geqslant \frac{2}{\pi}\varphi$ for $0 \leqslant \varphi \leqslant \frac{\pi}{2}$ we have

\begin{align} \int_0^{\pi} e^{-Rt\sin \varphi}\,d\varphi &= 2\int_0^{\pi/2} e^{-Rt\sin \varphi}\,d\varphi \\ &\leqslant 2\int_0^{\pi/2} \exp\bigl(-\tfrac{2Rt}{\pi}\varphi\bigr)\,d\varphi \\ &\leqslant 2 \int_0^{+\infty} \exp\bigl(-\tfrac{2Rt}{\pi}\varphi\bigr)\,d\varphi \\ &= \frac{\pi}{Rt}\,. \end{align}

Altogether the contribution by the circular arc is bounded by $\frac{C(t)}{R}$, and for fixed $t > 0$ this tends to $0$ as $R \to +\infty$, so we again obtain $(\ast)$. (Once again, the argument establishes only the existence as a principal value integral. Proving the existence as an improper Riemann integral using a circular arc is more cumbersome, that is then better done separately, using Dirichlet's criterion for example.)