Title: Part IB / Complex Methods
Tags: Contour_Integration Transform_Theory Laurent_Series_Singularities Möbius_Maps Harmonic_Functions Branch_Cut Conformal_Maps Analytic_Functions Calculus_of_Residues Laplaces_Equation

Q: Definition. The extended complex domain [...]
A: \(\mathbb{C}^* = \mathbb{C} \cup \{0\}\)
Back Extra: This is best seen by the Riemann sphere

Q: Definition. A function \(f(x,y)\) is harmonic if [...]
A: it satisfies \(\Delta f = \partial_x^2 f + \partial_y^2 f = 0\)

Q: Definition. Two functions u,v satisfying the Cauchy-Riemann equations are called [...]
A: harmonic conjugates
Back Extra: The C.R. equations relate real and imaginary parts of an analytic function. If we know one, we can determine the other up to a constant.

Q: Proposition. The real and imaginary parts of any analytic function are harmonic.
A: \(f(z) = u(x,y) + iv(x,y)\) analytic.
\(\partial_x (\partial_x u) = \partial_x(\partial_y v) = \partial_y (\partial_x v) = \partial_y (-\partial_y u) \implies \partial_x^2 u + \partial_y^2 u = 0\) and same for imaginary part.

Q: Definition. A branch point of a function \(f(z)\) is a point \(z_0\) that [...]
A: cannot be encircled by a curve C such that f is single-valued and continuous along C.
Back Extra: Also called a branch point singularity of f

Q: Proposition. Let \(f(z)\) have branch points \(z_1 , z_2, \dots \) A complete branch cut is a set of curves with
[...]
A: i) Every branch point has a cut ending on it
ii) Both ends of each cut end on a branch point
iii)Any curve in \(\mathbb{C}\) that does not intersect the branch cut either encloses all branch points or none of them
Back Extra: - Regard \(z=\infty\) as a branch point candidate
- The branch cut \( ((-\infty,-1) \cup (1, \infty))\) is really a single curve on the Riemann sphere across the North Pole.

Q: Definition. Möbius map. A map [...]
A: \(\mathcal{M} : \mathbb{C} \to \mathbb{C}, z \mapsto w = \frac{az + b}{cz+d}\) where \(a,b,c,d \in \mathbb{C}\) and \(ad \neq bc\).
Back Extra: If \(ad = bc\), \(\mathcal{M}\) maps all \(\mathbb{C}\) to a single point.
The map is analytic everywhere except at \(z = -d/c\)
Regarded as \(\mathcal{M} : \mathbb{C}* \to \mathbb{C}*\), \(\mathcal{M}\) is bijective with \(\mathcal{M}^{-1} : \mathbb{C}* \to \mathbb{C}*, w \mapsto z = \frac{-dw + b}{dw - a}\) also a Möbius map.

Q: Definition. A "circline" is [...]
A: either a circle or a line.

Q: Proposition. Any circline in \(\mathbb{C}\) is given by the points z with [...]
A: \[|z - z_1| = \lambda |z - z_2|\] with \(z_1 \neq z_2 \in \mathbb{C}, \lambda \in \mathbb{R}^+\)
Back Extra: Trivial for line, hard for circle.

Q: Proposition. A Möbius map maps a circline to a circline
A: Using the formula for \(\mathcal{M}^{-1}\), \(|z - z_1| = \lambda |z - z_2|\) \[ \left|-\frac{dw - b}{dw - a} - z_1 \right | = \lambda \left |- \frac{dw - b}{dw - a} - z_2 \right |\] then simplify and it works. If \(cz_1 + d = 0\) or \(cz_2 + d = 0\) this gives a circle. Otherwise \[ \left |w - \frac{az_1 + b}{cz_1 + d}\right| = \lambda \left|w - \frac{az_2 + b}{cz_2 + d}\right|\]

Q: Proposition. Let \(\alpha \neq \beta \neq \gamma \neq \alpha \in \mathbb{C}\) and \(\alpha' \neq \beta' \neq \gamma' \neq \alpha' \in \mathbb{C}\), then there exists a Möbius map that sends \(\alpha \to \alpha', \beta \to \beta', \gamma \to \gamma'\)
A: \[\mathcal{M}_1 (z) = \frac{\beta - \gamma}{\beta - \alpha} \frac{z - \alpha}{z - \gamma}\] sends \(\alpha \to 0, \beta \to 1, \gamma \to \infty\). \[\mathcal{M}_2(z) = \frac{\beta'  - \gamma'}{\beta' - \alpha'} \frac{z - \alpha'}{z - \gamma'}\] sends \(\alpha' \to 0, \beta' \to 1, \gamma' \to \infty\) implies \[\mathcal{M}_2^{-1} \circ \mathcal{M}_1\] is the desired map.

Q: Definition. \(f: U \to W; U,W \subseteq \mathbb{C}\) open, is a conformal map  iff [...]. If f is also bijective it is called a [...]
A: f is analytic with \(f'(z) \neq 0\) throughout U; conformal equivalence

Q: Proposition. A conformal map preserves the angle in magnitude and direction between two curves.
A: Let \(z_1(t)\) be a curve in \(\mathbb{C}\) with \(z_0 = z_1(t_0)\). \(\theta = \arg z_1'(t_0) = \) angle of the curve with the x-direction.
Let f be a conformal map, \(\implies \zeta_1(t) = f(z_1(t))\) is a new curve with tangent vector \[\zeta_1'(t_0) = \frac{df}{dz_1} \big |_{t = t_0} \frac{dz_1}{dt} \big |_{t = t_0} = f'(z_0) \cdot z_1'(t_0)\] angle with x-dir: \[\vartheta = \arg (\zeta_1'(t_0)) = \arg (f'(z_0) \cdot z_1'(t_0)) = \theta + \arg (f'(z_0))\] The rotation angle \(\arg (f'(z_0))\) is well-defined since \(f'(z_0) \neq 0\).
Let \(z_2(t)\) be a second curve through \(z_0\), it also gets rotated by \(arg(f'(z_0))\) and therefore the angle between \(z_1\) and \(z_2\) is unchanged.

Q: Proposition. When solving \(\Delta u\ = 0\) on U subject to Dirichlet boundary conditions on \(\partial U\), how is a solution on U given by \[\phi(x,y) = \Phi (\text{Re}(f(x + iy)), \text{Im}(f(x+iy)))\]
A: A solution \(\Phi(u,v)\) to the Laplace equation is harmonic, and hence, the real part of analytic complex function \(G(\zeta)\). The conformal map f is analytic by definition, so that \(g = G \circ f\) is a complex analytic function on U. Finally \(g(z) = G(f(z))\), so that the real part of g is given by the right hand side of the equation above, which tehrefore is a solution to Laplace's equation on U with the required boundary conditions.

Q: Definition. A curve \(\gamma\) is [...]
A: a continuous map \(\gamma : [0,1] \to \mathbb{C}\)
Back Extra: A curve has a direction from \(\gamma(0)\) to \(\gamma(1)\)

Q: Definition. A closed curve is [...]
A: a curve with \(\gamma(0) = \gamma(1)\)

Q: Definition. A simple curve is [...]
A: a curve that does not intersect itself except at the end points \(\gamma(0), \gamma(1)\)

Q: Definition. A contour is [...]
A: a piecewise differentiable curve

Q: Definition. The contour integral of a function f along a contour \(\gamma\) is \[[...]\]
A: \int_{\gamma} f(z) dz := \int_0^1 f(\gamma(t)) \gamma ' (t) dt
Back Extra: Alternatively dissect [0,1] into \(0 = t_0 < t_1 < \dots < t_n = 1\).
Let \(\delta t_k = t_{k+1} - t_k\), \(\delta z_k = z_{k+1} - z_k\) where \(z_k = \gamma(t_k)\)
Then \(\int_{\gamma} f(z) dz := \lim_{\Delta \to 0} \sum_{k=0}^{n-1} f(z_n)\delta z_n\) where \(\Delta := \max_{k = 0, \dots , n-1} \delta t_k\)

Q: Definition. An open set \(D \subseteq \mathbb{C}\) is a connected domain if [...] D is simply connected if [...]
A: for each pair \(z_1 , z_2 \in D\) can be connected by a curve whose image is also in \(D\).; it is connected and every curve in \(D\) encloses only points in \(D\)

Q: Theorem. If \(f(z)\) is analytic in a simply connected domain D and \(\gamma\) is a simple closed contour in D, then \[\oint_{\gamma} f(z)dz = 0\]
A: Green's theorem for functions P,Q with continuous partial derivatives on \(M \subseteq D\) where M is the interior of a simple closed contour \(\gamma\) in \(\mathbb{R}^2\). \[\oint_{\gamma} (Pdx + Qdy) = \iint_M (\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}) dx dy\] Write the complex \(f(z) = u(x,y) + iv(x,y)\), so \[\oint_{\gamma} f(z) dz = \oint_{\gamma} (u+iv)(dx + idy) = \oint (udx - vdy) + i \oint (vdx + udy) = \iint_M (-\frac{\partial v}{\partial x} - \frac{\partial u}{\partial y}) dx dy + i \iint_M (\frac{\partial u}{\partial x} - \frac{\partial v}{\partial y}) dx dy = 0\] using Cauchy-Riemann as f is analytic.

Q: Proposition. Let \(\gamma_1, \gamma_2\) be contours from a to b in \(\mathbb{C}\), and f(z) be analytic on \(\gamma_1\) and \(\gamma_2\) and the region bounded by them. Then \[\int_{\gamma_1} f(z) dz = \int_{\gamma_2} f(z) d(z)\]
A: Let \(\gamma_1, \gamma_2\) not intersect except at a,b. Then \(\gamma_1 - \gamma_2 := \gamma_1 + (-\gamma_2)\) is a simple closed contour which implies \[\oint_{\gamma_1 - \gamma_2} f(z) dz = 0\] by Cauchy's theorem. If \(\gamma_1 , \gamma_2\) intersect, dissect the curve at the intersection point and apply the proof to individual closed curves.

Q: Theorem. Cauchy's Integral Formula. Let \(f(z)\) be analytic on an open domain D, \(z_0 \in D\) and \(\gamma\) a simple closed contour in D encircling \(z_0\) counter clockwise. \[f(z_0) = \frac{1}{2\pi i} \oint \frac{f}{z - z_0} dz, \; f^{(n)}(z_0) = \frac{n!}{2\pi i} \frac{f(z)}{(z-z_0)^{n+1}} dz\]
A: Let \(\gamma_{\epsilon}\) be a circle of radius \(\epsilon\) around \(z_0\).
\(\frac{f(z)}{z-z_0}\) is analytic between \(\gamma\) and \(\gamma_{\epsilon}\).
Write \(z = z_0 + \epsilon e^{i \theta}, \theta \in [0,2\pi]\).
\[\oint_{\gamma} \frac{f(z)}{z-z_0} dz = \oint_{\gamma_{\epsilon}} \frac{f(z)}{z-z_0} dz = \int_0^{2\pi} \frac{f(z_0 + \epsilon e^{i \theta})}{\epsilon e^{i \theta}} i\epsilon e^{i \theta} d\theta = \lim_{\epsilon \to 0} i \int_0^{2\pi} f(z_0 + \epsilon e^{i \theta}) d\theta = 2\pi i f(z_0)\]
Take \(\frac{d}{dz_0}: f'(z_0)= \frac{1}{2\pi i} \oint_{\gamma} \frac{f(z)}{(z - z_0)^2} dz\) etc. to get the higher derivatives.

Q: Theorem. Liouville's Theorem. If f is analytic on \(\mathbb{C}\) na bounded, then \(f(z) =\) const.
A: \(\exists c_0 \in \mathbb{R}\) \(\forall z \in \mathbb{C}\) \(|f(z)| \leq c_0\). Let \(\gamma_r\) be a counter clockwise circle of radius r around \(z_0\). \[|f'(z_0)| = \left | \frac{1}{2\pi i} \oint_{\gamma_r} \frac{f(z)}{(z-z_0)^2} dz\right | \leq \frac{1}{2\pi} \int_{\gamma_r} \frac{c_0}{r^2} dz = \frac{c_0}{r} \to 0\] as \(r \to \infty\)

Q: Definition. The zeros of a function \(f(z)\) are the points where \(f(z_0) = 0\). A zero \(z_0\) is of order n if [...] A simple zero is [...]
A: in its Taylor expansion \(\sum_{k=0}^{\infty} a_k (z-z_0)^k\), the first non-zero coefficient is \(a_n\) or equivalently \(f(z_0) = f'(z_0) = \dots = f^{(n-1)}(z_0)\) but \(f^{(n)}(z_0) \neq 0\).; a zero of order 1.

Q: Definition. A singularity of \(f(z)\) is [...]
isolated singularity: [...]
non-isolated singularity: [...]
A: a \(z_0\) where f is not analytic.; f is analytic in a neighbourhood of \(z_0\)(but not at \(z_0\)); f is not analytic at \(z_0\) nor in any neighbourhood of \(z_0\)

Q: Proposition. Let \(f(z)\) have an essentially singularity at \(z_0\), then in any neighbourhood \(D\) of \(z_0\), \(f(z)\) takes on all possible complex values except at most 1.
A: No proof given.

Q: Definition. The residue, \(\text{Res}_{z = z_0} f(z)\) of a function f with isolated singularity \(z = z_0\) is [...]
A: the coefficient \(a_{-1}\) of the Laurent series of f about \(z_0\).

Q: Proposition. Let f have a pole of order n at \(z = z_0\), then \[\text{Res}_{z = z_0} f(z) = \lim_{z \to z_0} \frac{1}{(n-1)!} \frac{d^{n-1}}{dz^{n-1}} [(z-z_0)^n f(z)]\] and for n = 1 \[\text{Res}_{z = z_0} f(z) = \lim_{z \to z_0} [(z-z_0)f(z)]\]
A: For n = 1, the Laurent series starts \(f(z) = a_{-1}\frac{1}{z-z_0} + a_0 + \dots \implies \lim_{z \to z_0}[(z-z_0)f(z)] = a_{-1}\) and then use induction for n > 1.

Q: Theorem. Let \(\gamma\) be a simple closed contour traversed in counter-clockwise direction and f(z) be analytic \(\gamma\) except for an isolated singularity \(z_0\). Then \[\oint_{\gamma}f(z) dz = i2\pi a_{-1} = i2\pi \text{Res}_{z=z_0} f(z)\]
A: f(z) is analytic except at \(z_0\). Deform the contour \(\gamma\) into a circle \(\gamma_r \) inside \(\gamma\) around \(z_0\)
\[\oint_{\gamma}f(z)dz = \oint_{\gamma_r}f(z)dz = \oint_{\gamma_r} \sum_{n = -\infty}^{n = \infty} a_n(z-z_0)^n dz = \sum_{n=-\infty}^{\infty} a_n \oint_{\gamma_r} (z-z_0)^n dz\] \[\oint_{\gamma_r}(z-z_0)^n d = \oint_0^{2\pi} r^n e^{in\theta}ire^{i\theta} d\theta = ir^{n+1}\oint_{0}^{2\pi} e^{i(n+1)\theta} d\theta = \begin{cases} i2\pi & \text{ for } n =1 \\ \frac{r^{n+1}}{n+1} [e^{i(n+1)\theta}]_0^{2\pi} = 0  & \text{ for } n \neq -1 \end{cases}\]
so only the \(a_{-1}\) term survives.

Q: Theorem. Residue Theorem. Let \(f(z)\) be analytic in a simply connected domain \(\mathcal{D}\) except at a finite number of isolated singularities \(z_1, \dots , z_n\). Let \(\gamma\) be a simple closed counter-clockwise contour in \(\mathcal{D}\) that encircles all \(z_k, k = 1, \dots , n\). Then \[\oint_{\gamma} f(z)dz = i2\pi \sum_{k=1}^n \text{Res}_{z = z_k}f(z)\]
A: Approximate \(\gamma\) with \(\hat{\gamma}\) with the \(z_k\) cut out.
\(\hat{\gamma}\) does not encircle any singularities now so \[0 = \oint_{\hat{\gamma}} f(z)dz = \oint_{\gamma}f(z)dz + \sum_{k=1}^n \oint_{\gamma_k} f(z)dz\] \[\implies \oint_{\gamma}f(z) = - \sum_{k=1}^n \oint_{\gamma_k} f(z)dz = i2\pi \sum_{k=1}^n \text{Res}_{z = z_k} f(z)\]

Q: Lemma.  Jordan's Lemma. Let \(f(z)\) be analytic in \(\mathbb{C}\), except for a finite number of singular with \(f(z) \to 0\) as \(|z| \to \infty\). Let \(\lambda, \mu \in \mathbb{R}, \lambda > 0, \mu < 0\). Then \[\lim_{R \to \infty} \int_{\gamma_R} f(z)e^{i \lambda z} dz = 0\] \[\lim_{R \to \infty} \int_{\overline{\gamma}_R} f(z)e^{i \mu z} dz = 0\]
A: One can show that \[\sin x \geq \frac{2}{\pi} x\] for \(x \in [0, \pi / 2]\). Parametrise: \(\gamma_R : \theta \mapsto R e^{i \theta}\) \[\left | \int_{\gamma_R} f(z) e^{i \lambda z} dz\right | = \left | \int_0^{\pi} f(Re^{i \theta}) e^{i \lambda e^{i \theta}} i R e^{i \theta} d\theta \right |\] \[\leq R \int_0^{\pi} |f(Re^{i\theta})| \cdot |e^{i \lambda R e^{i \theta}} | d\theta \] \[ = R\int_0^{\pi} |f(Re^{i \theta})| \cdot e^{- \lambda R \sin \theta} d\theta \leq R \sup_{z \in \gamma_R} |f(z)| \cdot 2 \int_0^{\pi / 2} e^{- \lambda R \sin \theta} d\theta\] \[\leq 2R \sup_{z \in \gamma_R} |f(z)| \int_0^{\pi / 2} e^{ - 2 \lambda R \theta / \pi} d\theta = \frac{\pi}{\lambda} (1 - e^{-\lambda R}) \sup_{z \in \gamma_R} |f(z)| \to 0\] as \(R \to \infty\) and likewise for \(\mu < 0\) along \(\overline{\gamma}_R\)

Q: Definition. Let \(f : \mathbb{R} \to \mathbb{C}\) be absolutely integrable, i.e. \(\int_{-\infty}^{\infty} |f(x)|dx\) exists, have bounded variation and a finite number of discontinuities. Then its Fourier Transform and its inverse are defined as [...]
A: \[\tilde{f}(k) = \mathcal{F}[f(x)](k) = \int_{-\infty}^{\infty}f(x)e^{-ikx}dx\] \[f(x) = \mathcal{F}^{-1}[\tilde{f}(k)](x) = \frac{1}{2\pi}\int_{-\infty}^{\infty} \tilde{f}(k)e^{ikx} dk\]

Definition. The Cauchy principal value of an integral \(\int_{-\infty}^{\infty}g(x)dx\) is defined as

Q: Definition. Let \(f(t)\) be a function defined for all \(t \geq 0\). The Laplace transform of \(f(t)\) is given by [...]
A: the integral \[F(s) = \mathcal{L}\{f(t)\} (s) := \int_0^{\infty}f(t)e^{-st}dt \; \; s \in \mathbb{C}\] provided that the integral exists
Back Extra: Works for functions that grow no faster than exponential

Proposition.
1. Linearity: For constants \(\alpha, \beta \in \mathbb{C}\) \[\mathcal{L} \{\alpha f + \beta g\} = \alpha \mathcal{L}\{f\} + \beta \mathcal{L} \{g\} \]
2. Translation: For a constant \(t_0 \in \mathbb{R}\), \[\mathcal{L}\{f(t-t_0) H(t- t_0) \} (s) = e^{-st_0}F(s)\] where H denotes the Heaviside step function.
3. Scaling: For a constant \(\lambda > 0\), \[\mathcal{L}\{f(\lambda t)\}(s) = \frac{1}{\lambda} F(\frac{s}{\lambda})\]
4. Shifting: For a constant \(s_0 \in \mathbb{C}\), \[\mathcal{L}\{e^{s_0 t} f(t)\}(s) = F(s - s_0)\]
5. Transform of a derivative: \[\mathcal{L}\{f'(t)\}(s) = sF(s) - f(0)\]
6. Derivative of a transform: \[F'(s) = \mathcal{L}\{-tf(t)\}(s)\] More generally, we have \[F^{(n)}(s) = \mathcal{L}\{(-t)^n f(t)\}(s)\]
7. Asymptotic limits: \[\lim_{s \to \infty} sF(s) = f(0)\] \[\lim_{s \to 0} sF(s) = f(\infty)\] provided the limit \(\lim_{t \to \infty} f(t)\) exists.

Q: Proposition. Given a function \(F(s)\), we can compute its inverse Laplace transform \(f(t)\) from the Bromwich inversion formula \[f(t) = \frac{1}{i2 \pi} \int_{\alpha - i\infty}^{\alpha + i \infty} F(s)e^{st}ds\] where \(\alpha\) is a real constant chosen such that the Bromwich inversion contour \(\gamma\) consisting of the point \(\{s \in \mathbb{C} : \text{Re}(s) = \alpha\}\) lies to the right of all singularities of F(s).
A: Since f(t) has a Laplace transform, we have \(f(t) = 0\) for \(t < 0\) and f does not grow faster than exponential. We can therefore choose an \(\alpha \in \mathbb{R}\) such that \[g(t) = f(t)e^{-\alpha t}\] decays exponentially as \(t \to \infty\) and therefore has a Fourier transform, \[\tilde{g}(\omega) = \int_{-\infty}^{\infty} f(t) e^{- \alpha t} e^{-i \omega t} dt = F(a + i \omega)\] This enables us to apply the inverse Fourier transform, whence \[g(t) = \frac{1}{2\pi} \int_{- \infty}^{\infty}F(\alpha + i \omega) e^{i \omega t} d\omega\]
Finally we substitute \(s = \alpha + i \omega \implies d\omega = ds / i\), and obtain \[f(t)e^{- \alpha t} = \frac{1}{2 \pi i} \int_{a - i \infty}^{a + i\infty} F(s)e^{(s-\alpha)t} ds\] \[\implies f(t) = \frac{1}{i2\pi} \int_{a - i \infty}^{a + i \infty} F(s)e^{st}ds\] This additional requirement that the contour of points with \(\text{Re}(s) > \alpha\) lies to the right of all singularities of F(s) fixed a constant of integration and thus ensures that \(f(t) = 0\) for \(t < 0\); we will see how this works in the next proof.

Q: Proposition. Let F(s) be the Laplace transform of a function f(t) and have only a finite number of isolated singularities at points \(s_k \in \mathbb{C}\), \(k = 1, \dots, n\). Let furthermore \(F(s) \to 0\) as \(|s| \to \infty\). Then \(f(t) = 0\) for \(t < 0\) and for \(t > 0\), \[f(t) = \sum_{k=1}^n \text{Res}_{s = s_k} (F(s) e^{st})\]
A: First let us consider the case \(t<0\) and construct the contour \(\gamma = \gamma_0 + \overline{\gamma}_r \) from our last diagram. If \(F(s) = o(s^{-1})\), we directly obtain \[\left | \int_{\overline{\gamma}_R} F(s) e^{st} ds\right | \leq \pi R e^{\alpha t} \sup_{s \in \overline{\gamma}_R} |F(s)| \to^{R \to \infty} 0\]
Note that we have used here that \(\alpha \leq \text{Re}(s)\) along \(\overline{\gamma}_R\) and for \(t < 0\) we therefore get \(\alpha t \geq \text{Re}(s) t\) and \(|e^{\alpha t}| \geq |e^{st}|\) along \(\overline{\gamma}_R\). One can show, using a modified version of Jordan's lemma, that this result still holds if F(s) approaches less rapidly that \(o(s^{-1})\). We skip the details and merely note that in any case, \[\lim_{R \to \infty} \int_{\overline{\gamma}_R} F(s)e^{st}ds = 0\] and therefore \(\int_{\gamma_0} F(s)e^{st}ds = \int_{\gamma_0 + \gamma_R} F(s)e^{st}ds\) in the limit \(R \to \infty\). But out contour does not enclose any singularities of the integrand, so that by Cauchy's Theorem, we get \(f(t) = 0\) for \(t < 0\) as expected.
For \(t > 0\), we close the contour to the left of \(\text{Re}(s) = \alpha\) as shown in the diagram. Since \(F(s)\) only has a finite number of isolated singularities, we will encircle the whole lot as \(R \to \infty\). As before, one shows that the contribution along \(\gamma_R\) vanishes as \(R \to \infty\). We can then use the residue theorem to compute \(f(t)\) according to the Bromwich inversion formula, \[f(t) = \frac{1}{i2 \pi} \lim_{R \to \infty} \int_{\gamma_0} F(s)e^{st}ds = \frac{1}{i2\pi} \lim_{R\to \infty} \int_{\gamma_0 + \gamma_R} F(s)e^{st} ds = \sum_{k=1}^n \text{Res}_{s = s_k} (F(s)e^{st})\]

Q: Theorem. The Laplace transform of a convolution is given by \[\mathcal{L}\{f * g\}(s) = \mathcal{L}\{f\}(s) \cdot \mathcal{L}\{g\}(s) = F(s)G(s)\]
A: \[\mathcal{L}\{f * g \}(s) = \int_0^{\infty} \left [ \int_0^t f(t-u)g(u) du \right ] e^{-st}dt = \int_0^{\infty} \left [ \int_u^{\infty} f(t-u)g(u) e^{-st} dt\right ] du\] where we have swapped the order of integration and adjusted the limits such that the integration area remains the same. Defining \(x = t-u\), we obtain \[\mathcal{L}\{ f * g \}(s) = \int_0^{infty} \left [ \int_0^{\infty} f(x)g(u)e^{-sx} e^{-su} dx\right ]du\] \[= \int_0^{\infty} \left [ \int_0^{\infty} f(x)e^{-sx} dx\right ]g(u)e^{-su} du = F(s)G(s)\]