I'm trying to learn how to evaluate inverse Laplace transforms without the aid of a table of transforms, and I've found the inversion formula: $$\mathcal{L}^{-1}\{F\}(t)=\frac{1}{2\pi i}\int_{\gamma-i\infty}^{\gamma+i\infty}F(s)e^{st}ds$$ I'm currently in high school, and I don't a lot of knowledge in terms of complex analysis, but I do have a pretty good base in calculus, so if anyone could help me out, it'd be great if you could also explain some of the complex analysis methods that may be involved in evaluating this integral. A step by step solution for the ILT of $F(s)=\frac{1}{s}$ would be awesome. Thanks in advance.
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$\begingroup$Before I start, it would be very hard to do this kind of transform without learning how to evaluate contour integrals, how to evaluate residues (it gets more complicated for something like $\frac{1}{z^3})$ (mentioned in the comments), and Jordan's Lemma to justify the integral along the circular part to go to $0$.
However this question might be useful to someone who knows a little more about complex analysis, so I'll run through the evaluation of the inverse Laplace transform of $F(s)=\dfrac{1}{s}$ as an example.
As you have seen, the integral in question is $\frac{1}{2\pi i}\int_{\gamma-i\infty}^{\gamma+i\infty}F(s)e^{st}ds$. What do the limits mean?
Firstly to determine $\gamma$, you need to examine the function $F(s)e^{st}$, and more specifically, where its singularities lie. In this example, there is one singularity, at $s=0$. You choose $\gamma$ to be a real number which lies to the right of all the singularities of your function when sketched in the complex plane. Here, since our singularity is at $0$, we can choose $\gamma$ to be, say $1$. Then the $\pm i\infty$ simply mean to express the contour going up to $\infty$ and down to $-\infty$ (though this is technically also $\infty$) from the point $\gamma$ on the complex plane. So far we have described a contour which is just a vertical line, which has equation $Re(z)=\gamma$.
Now the question is, how do we evaluate this integral? This is where the complex analysis methods come in. We first draw a contour as follows:
The ends $A$ and $B$ are at the points $\gamma-iR$ and $\gamma+iR$ respectively. Taking the limit as $R\rightarrow\infty$, the line $AB$ becomes the integral we want. Meanwhile the part of the circle which goes from $B$ to $C$ all the way round to $A$ becomes a very big circle when we take this limit. There is a theorem, called Jordan's Lemma, which can be used to show that the integral along this contour goes to $0$ in specific cases, based on what $F(s)$ is. In this example, it does work. So as we take the limit, the integral along this contour is exactly the same as the integral we wish to calculate.
Next - how do we evaluate the integral along this contour? This requires the residue theorem. In short, you evaluate something called a residue at each singularity inside the contour. Here, that residue is at $s=0$. The sum of all the residues inside a contour is equal to the integral along the contour, divided by $2\pi i$, just as we want. So to work out this integral you need to work out the residues at each singularity, and add them up.
In your example, $F(s)=\frac{1}{s}$, so $$Res_{s=0}(F(s))=\lim_{s\rightarrow0} \underbrace{\frac{e^{st}}{s}}_{\text{the integral}}\times (s-\underbrace{z_0}_{\text{where the singularity is}})=\lim_{s\rightarrow0}(e^{st})=1.$$
So we conclude that this is the inverse Laplace transform of $F(s)=\frac{1}{s}$. $$\frac{1}{2\pi i}\int_{\gamma-i\infty}^{\gamma+i\infty}F(s)e^{st}ds=1$$
$\endgroup$ 11 $\begingroup$A complex integral can be defined via the path taken by the contour parameterized by a real variable, that is for $t\mapsto z(t)$, $ a \leq t \leq b$ $$ \int_C f(z)dz = \int_a^b f(z(t))z'(t)dt. $$ In the inverse Laplace transform the parameterization is $t \mapsto \gamma+it$ and $-\infty \leq t \leq \infty$. We thus have $$ L^{-1}(F(s)) = \frac{1}{2\pi}\int_{-\infty}^\infty F(\gamma+it)e^{\gamma+it}dt. $$ Generally we take $\gamma = 0$ but since your example has a pole at the origin you will need to take a different approach. As chappers mentioned this would involve a lengthy discussion of the concepts in analytic function theory.
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