I was reading this question (Inverse Laplace Transform of $1/(s+1)$ without table) and read the following:
The pole is on the left half plane, so: $$\gamma(t)=\frac{1}{2i\pi}\int ^{i\infty}_{-i\infty}\frac {e^{st}}{s+1}\,\mathrm{d}s$$
What exactly is a "pole" of a function? Based on this other question I was reading (Finding poles, indicating their order and then computing their residues), it seems that the "pole" of a function is at all points where you are forced to divide by 0. Is this correct?
Once you have determined the "pole" of a function - how do you determine that the "pole" of the function 1/(s+1) is on the left half of the plane? Is this done by determining if the "pole" of this function has a "complex part"? (e.g. a + ib)
Thank you!
Note: I am asking this question with regards to calculating Inverse Laplace Transforms. For many common mathematical functions, I am guessing that the "poles" of these functions will be in the "left plane" - and thus, the "Gamma" term in the Inverse Laplace Transform will end up being 0 and thus facilitate the computation of the integral required in the Inverse Laplace Transform. Is this correct?
The point $z_{0}$ is an isolated singularity of $f(z):\mathbb{C}\mapsto\mathbb{C}$ if $f(z)$ is analytic in $0 < |z-z_{0}| < r$, which is a punctured disk of radius $r$ centered at $z_{0}$ but not analytic at $z_0$. A pole is a type of isolated singularity.
A complex function $f(z)$ has a pole of order $m$ at a point $z_{0}$ if $m$ is the largest positive integer such that $a_{-m} \ne 0$ where $a_n$ is the coefficient in front of the term with power $n$ in $f(z)$’s Laurent series. A pole of order one is a simple pole. A pole of order two is a double pole, etc. The limit to a pole “exists” and is infinite.
Intuitively, you can think of a pole as a vertical asymptote in the complex plane. It turns out that when you integrate along a closed contour in the complex plane, the path you travel is only dependent on the residues of the poles inside the contour. This is known as the residue theorem.
We calculate the residue of a function at a pole through either the limit formula or, if one expands a function $f(z)$ in a Laurent series about the point $z_{0}$, we can look at the coefficient with power $-1$ to find the residue.
Example: $$f(z) = \frac{1}{(2z-5i)^{69}}$$ has a pole of order 69 at $z=\frac{5i}{2}$.
Well, for the inverse Laplace transform, we use a Bromwich contour, which is a straight line on the positive real axis and what usually is a large semicircle that connects the ends of the line. We determine how far on the positive real axis we need to set the straight line based on the poles of $f(z)$, such that when we draw the contour it encompasses every single pole.
Sometimes you might encounter a branch cut so you deform your path around it like this
Hopefully this helped. I recommend watching YouTube videos or reading Wikipedia/textbooks on complex analysis though before you try anything further.