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)
 A: 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.)
