Questions

0. Prove that for any holomorphic functions \(F: \Omega \to \mathbb{C}^{m\times n}\) and \(G:\Omega \to \mathbb{C}^{n \times p}\), the matrix function product \[FG:z \in \Omega \mapsto F(z) G(z) \in \mathbb{C}^{m \times p}\] is holomorphic and \((FG)' = F' G + F G'\).

Let \(p \in \mathbb{N}^*\) and \(A \in \mathbb{C}^{p \times p}\). We denote \(\Omega\) the set of complex numbers \(z\) such that the matrix \(I - zA\) is invertible and \(F\) the matrix-valued function defined by \[ F: z \in \Omega \mapsto [I - zA]^{-1}. \]

1. Show that \(\Omega\) is open. Relate the set \(\mathbb{C} \setminus \Omega\) of singularities of \(F\) and the spectrum of \(A\).

2. Show that the function \(F\) is holomorphic and compute its derivative.

   Hint: define the difference quotient \[ \epsilon(z, h) = \frac{F(z+h) - F(z)}{h} \] then compute \(\epsilon(z,h) [I - z A] [I - (z+h) A]\) (optionally, consider first the case with \(p=1\) and \(A = [a]\) for some \(a \in \mathbb{C}\) to guess the expression of the derivative in the general case).

3. Prove that the \(n\)-th order derivative \(F^{(n)}\) of \(F\) exists for every natural number \(n\) and satisfies \[ \forall \, z \in \Omega, \; F^{(n)}(z) = n! A^n [I-zA]^{-n-1}. \] Hint: prove the result by induction on \(n\).

4. What is the (possibly infinite) radius \(r(A)\) of the largest open disk centered on the origin and included in \(\Omega\) ? Relate \(r(A)\) and the spectral radius of \(A\), defined as \[ \rho(A) = \max \, \{|\lambda| \; | \; \lambda \in \sigma(A)\}. \] Find a sequence of matrices \(A_n \in \mathbb{C}^{p \times p}\) such that \[ \forall \, z \in D(0,r(A)), \; F(z) = [I - zA]^{-1} = \sum_{n=0}^{+\infty} A_n z^n. \] (the convergence of this matrix series should be interpreted element-wise).

5. Let \(n \in \mathbb{N}\). Let \(z \in \mathbb{C} \setminus \sigma(A)\) such that \(z \neq 0\); write \(z^n[z I - A]^{-1}\) as a function of \(z\) and \(F(z^{-1})\). Show that the function \[z \in \mathbb{C} \setminus \sigma(A)\mapsto z^n [zI - A]^{-1}\] is complex-differentiable.

6. Compute the line integral \[ \frac{1}{i2\pi}\int_{\gamma} z^n [zI - A]^{-1} dz \] first when \(\gamma = r[\circlearrowleft]\) with \(r > \rho(A)\), then more generally for any sequence of rectifiable closed paths \(\gamma\) of \(\mathbb{C} \setminus \sigma(A)\) such that \(\forall \, \lambda \in \sigma(A)\), \(\mathrm{ind}(\gamma, \lambda) = 1\).

7. Let \[ A = \left[ \begin{array}{rr} 0 & -1 \\ 1 & 0 \end{array} \right] \] and let \(\gamma\) be a sequence of rectifiable closed paths of \(\mathbb{C} \setminus \mathbb{R}_-\) such that \(\forall \, \lambda \in \sigma(A)\), \(\mathrm{ind}(\gamma, \lambda) = 1\). Compute \([zI - A]^{-1}\) and then the matrix \(\log A\), defined as \[ \log A = \frac{1}{i2\pi} \int_{\gamma} (\log z) [z I - A]^{-1} \, dz. \]

Answers

0. For any \(z \in \Omega\), \(j \in \{1,\dots,m\}\) and \(l \in \{1,\dots, p\}\), \[ FG(z)_{jl} = \sum_{k=1}^{n} F(z)_{jk} G(z)_{kl} = \sum_{k=1}^n f_{jk}(z) g_{kl}(z). \] By the sum and product rules for complex-valued holomorphic functions, every element of \(FG\) is holomorphic and \[ \begin{split} (FG)'(z)_{jl} &= \sum_{k=1}^n f'_{jk}(z) g_{kl}(z) + \sum_{k=1}^n f_{jk}(z) g'_{kl}(z) \\ &= F'G(z)_{jl} + F G'(z)_{jl}. \end{split} \]

1. The function \(d: z \in \mathbb{C} \mapsto \det (I - zA)\) is a polynomial, thus it is continuous; the matrix \(I - z A\) is invertible if and only if \(d(z) \neq 0\), hence \(\Omega = d^{-1}(\mathbb{C}^*)\) is open as the preimage of an open set by a continuous function.

   Since the matrix \(I - z A\) is invertible for \(z=0\), we have \(0 \in \Omega\); if \(z\neq 0\), \[I - z A = z (z^{-1}I - A)\] and it is invertible if and only if \(z^{-1}\) is not an eigenvalue of \(A\). Finally, \[ \mathbb{C} \setminus \Omega = \{\lambda^{-1} \; | \; \lambda \in \sigma(A) \setminus \{0\}\}. \]

2. For any pair \((k,l) \in \{1,\dots,p\}^2\), any \(z\in\Omega\) and \(h \in \mathbb{C}\) small enough, \[ \epsilon(z,h)_{kl} = \frac{f_{kl}(z+h) - f_{kl}(z)}{h} \] is the difference quotient of \(f_{kl}\) at \(z\). Therefore, \(F\) is complex-differentiable at \(z\) if (and only if) \(\epsilon(z, h)\) has a limit element-wise when \(h \to 0\) and \(F'(z)\) is this limit. Now, we have \[ \begin{split} \epsilon(z,h) & = \frac{[I - (z+h) A]^{-1} - [I - zA]^{-1}}{h}\\ \end{split} \] The matrices \(I - (z+h)A\) and \(I - zA\) commute, hence \[ \epsilon(z,h) [I -(z+h) A][I - zA] = \frac{[I - zA] - [I - (z+h)A]}{h} = A, \] thus \(\epsilon(z, h) = A [I - z A]^{-1} [I - (z+h)A]^{-1}\). Since matrix inversion is continuous \[ F'(z) = \lim_{h \to 0} \epsilon(z,h) = A [I - z A]^{-2}. \]

3. The function \(F\) is holomorphic; all of its elements are holomorphic, thus they have derivatives at all orders and so does \(F\). Assume that the \(n+1\)-th order derivative of \(F\) satisfies \[ F^{(n+1)}(z) = n A [I-zA]^{-1} F^{(n)}(z). \] This property certainly holds for \(n=0\), as \(F'(z) = A [I-zA]^{-1} F(z)\). Now, by the product rule, as \(A\) and \(F(z) = [I - zA]^{-1}\) commute, \[ \begin{split} F^{(n+2)}(z) &= n A F'(z) F^{(n)}(z) + n A F(z) F^{(n+1)}(z) \\ &= A F(z) (n A F(z) F^{(n)}(z)) + n A F(z) F^{(n+1)}(z) \\ &= A F(z) F^{(n+1)}(z) + n A F(z) F^{(n+1)}(z) \\ &= (n+1) A F(z) F^{(n+1)}(z) \\ \end{split} \] Consequently, the formula holds for all \(n\), which leads to \[ F^{(n)}(z) = n! A^n [I-zA]^{-n-1}. \]

4. The radius \(r(A)\) of the largest disk centered on the origin and included in \(\Omega\) is \[ r(A) = \min \{|\lambda^{-1}| \; | \; \lambda \in \sigma(A) \setminus \{0\}\} \\ \] if \(\sigma(A) \setminus \{0\}\) is non-empty and \(r(A) = +\infty\) otherwise. If we adopt the convention \(0^{-1} = +\infty\), we can rewrite this as \[ r(A) = (\max \{|\lambda| \; | \; \lambda \in \sigma(A)\})^{-1} = \rho(A)^{-1}. \] The function \(F\) is holomorphic in \(\Omega\), thus every element \(f_{kl}\) is holomorphic in \(\Omega\), which has a Taylor series expansion in \(D(0,r)\): \[ \forall \, z \in D(0,r), \; f_{kl}(z) = \sum_{n=0}^{+\infty} \frac{f^{(n)}_{kl}(0)}{n!} z^n. \] Consequently, \[ \forall \, z \in D(0,r), \; F(z) = \sum_{n=0}^{+\infty} \frac{F^{(n)}(0)}{n!} z^n = \sum_{n=0}^{+\infty} A^n z^n. \]

5. If \(z \in \mathbb{C} \setminus \sigma(A)\) and \(z\neq 0\), then \[ z^n[z I - A]^{-1} = z^{n-1}[I - z^{-1}A]^{-1} = z^{n-1} F(z^{-1}). \] Consequently, \(z \in \mathbb{C} \setminus \sigma(A) \mapsto z^n[z I - A]^{-1}\) is complex-differentiable at \(z\) – by the product and chain rules – if \(z\neq 0\), which is all we need unless \(0 \not \in \sigma(A)\). In this case, \(A\) is invertible and by continuity of the matrix inversion, the function \(z\mapsto z^n[zI-A]^{-1}\) is continuous at \(z=0\). Hence, \(0\) is a removable singularity of \(z\mapsto z^{n-1}F(z^{-1})\) and the function \(z \in \mathbb{C} \setminus \sigma(A) \mapsto z^n[zI-A]^{-1}\) is also complex-differentiable at \(z=0\).

6. If \(|z|>\rho(A)\), \(z \in \mathbb{C} \setminus \sigma(A)\), \(|z^{-1}| < r(A)\) and \[ z^n[z I - A]^{-1} = z^{n-1}[I - z^{-1}A]^{-1} = z^{n-1} \sum_{l=0}^{+\infty} A^{l} z^{-l} = \sum_{l=0}^{+\infty} A^{l} z^{n-1-l}. \] This convergence if uniform (element-wise) on compact subsets of the annulus, therefore \[ \begin{split} \frac{1}{i2\pi} \int_{r[\circlearrowleft]} z^n [zI - A]^{-1} dz &= \frac{1}{i2\pi} \int_{r[\circlearrowleft]} \sum_{l=0}^{+\infty} A^{l} z^{n-1-l} dz \\ &= \sum_{l=0}^{+\infty} A^{l} \left[\frac{1}{i2\pi} \int_{r[\circlearrowleft]} z^{n-1-l} dz\right] \\ \end{split} \] The integral \(\int_{r[\circlearrowleft]} z^{p} \, dz\) is zero unless the integer \(p\) is \(-1\), in which case it is equal to \(i2\pi\), hence \[ \frac{1}{i2\pi} \int_{r[\circlearrowleft]} z^n [zI - A]^{-1} dz = A^{n} \]

   Now, If \(\gamma\) is a sequence of rectifiable closed paths of \(\mathbb{C} \setminus \sigma(A)\) such that \(\forall \, \lambda \in \sigma(A)\), \(\mathrm{ind}(\gamma, \lambda) = 1\) and \(r > \rho(A)\), the sequence of paths \(\mu =r[\circlearrowleft] \, | \, \gamma^{\leftarrow}\) satisfies \(\forall \, z \in \sigma(A)\), \(\mathrm{ind}(\mu,z) = 0\). Cauchy’s integral theorem provides \(\int_{\mu} z^n[zI-A]^{-1} \, dz = 0\), or equivalently, \[ \frac{1}{i2\pi} \int_{\gamma} z^n[zI-A]^{-1} \, dz = \frac{1}{i2\pi} \int_{r[\circlearrowleft]} z^n[zI-A]^{-1} \, dz = A^n. \]

7. We have \[ z I - A = \left[ \begin{array}{rr} z & +1 \\ -1 & z \end{array} \right] \] hence \(\det (zI -A)= z^2 +1\). The spectrum of \(A\) is \(\sigma(A) = \{-i, +i\}\) and \[ [zI - A]^{-1} = \frac{1}{z^2+1} \left[ \begin{array}{rr} z & -1 \\ +1 & z \end{array} \right]. \] Consequently, \[ \frac{1}{i2\pi} \int_{\gamma} (\log z) [z I - A]^{-1} \, dz = \frac{1}{i2\pi} \int_{\gamma} \left[ \begin{array}{rr} f(z) & -g(z) \\ g(z) & f(z) \end{array} \right]\, dz \] with \[ f(z) = \frac{z\log z}{z^2+1}, \; g(z) = \frac{\log z}{z^2+1}. \] Let \(\Omega = \mathbb{C} \setminus \mathbb{R}_- \setminus \sigma(A)\); this set is open. Both functions \(f\) and \(g\) are holomorphic on \(\Omega\), have isolated singularities in \(\sigma(A)\) and are the quotient of holomorphic functions \(a\) and \(b\) defined on \(\Omega \cup \sigma(A)\) such that when \(\lambda \in \sigma(A)\), \(a(\lambda) \neq 0\), \(b(\lambda) = 0\) and \(b'(\lambda) \neq 0\). Thus, we have \[ \sum_{z \in \sigma(A)} \mathrm{ind}(\gamma, z) \mathrm{res}(f, z) = \frac{(-i)\log (-i)}{2(-i)} + \frac{i\log i}{2i} = 0 \] and \[ \sum_{z \in \sigma(A)} \mathrm{ind}(\gamma, z) \mathrm{res}(g, z) = \frac{\log (-i)}{2(-i)} + \frac{\log i}{2i} = \frac{\pi}{2}. \] Since the open set \[ \Omega \cup \sigma(A) = \mathbb{C} \setminus \mathbb{R}_- \] is simply connected (it is star-shaped), \(\mathrm{Int} \, \gamma \subset \Omega \cup \sigma(A)\). Finally, the residue theorem provides \[ \log A = \frac{1}{i2\pi} \int_{\gamma} (\log z) [z I - A]^{-1} \, dz = \frac{\pi}{2} \left[ \begin{array}{rr} 0 & -1 \\ +1 & 0 \end{array} \right] \]

Questions

We suppose that \(f\) is complex-differentiable on \(\mathbb{C}\) and denote \[ f(z) = \sum_{n=0}^{+\infty} a_n z^n \] its Taylor expansion at \(0\).

1. Show that if \(f\) is of exponential type, \(\sigma(f) \geq 0\) unless \(f\) is identically zero.

2. Show that the function \[ f_0: z \in \mathbb{C} \to e^{-z} \] is of exponential type and compute \(\sigma(f_0)\).

3. Prove that if for some \(\kappa > 0\) and \(\sigma \geq 0\), \(\forall \, n \in \mathbb{N}\), \(|n! a_n| \leq \kappa \sigma^n\), then \(f\) is of exponential type and \(\sigma(f) \leq \sigma\).

4. Prove that if \(\forall \, z \in \mathbb{C}, \; |f(z)| \leq \kappa e^{\sigma |z|}\) then \[ \forall \, n \in \mathbb{N}, \; \forall \, r > 0, \; |a_n| \leq \kappa \frac{e^{\sigma r}}{r^n}. \] Let \(n \in \mathbb{N}\); what is the value of \(r\) that provides the tightest upper bound on \(|a_n|\)? Compute this bound.

5. Prove that if \(f\) is of exponential type and \(\sigma > \sigma(f)\), there is a \(\kappa > 0\) such that \(\forall \, n \in \mathbb{N}\), \(|n! a_n| \leq \kappa \sigma^n\). Reminder – Stirling’s formula: \[ n! \sim \sqrt{2\pi n} \left(\frac{n}{e}\right)^n, \; \mbox{ that is } \; \lim_{n \to + \infty} \frac{n!e^n}{\sqrt{2\pi n} n^n} = 1. \]

6. Suppose that \(f\) is of exponential type; show that the function \(g:\mathbb{C} \to \mathbb{C}\) defined by \[ g(z) = \sum_{n=0}^{+\infty} |a_n| z^n \] is of exponential type and \(\sigma(g) = \sigma(f)\).

7. What is the domain of the Laplace transform of the function \(f_0\)? Compute \(\mathcal{L}[f_0](s)\) for any \(s\) in this domain. Prove that in general the domain of the Laplace transform of a function of exponential type \(f\) contains the set \(\{s \in \mathbb{C} \; | \; \mathrm{Re} \, \, s > \sigma(f)\}\).

8. Show that whenever \(\mathrm{Re} \, \, s >\sigma(f)\), \[ \mathcal{L}[f](s) = \int_{\mathbb{R}_+} f(t) e^{-st} \, dt = \sum_{n=0}^{+\infty} \frac{n! a_n}{s^{n+1}}. \] Prove that \(\mathcal{L}[f]\) has a unique holomorphic extension to the open set \[\Omega = \mathrm{dom} \, \mathcal{L}[f] \cup \{s\in \mathbb{C} \; | \; |s| > \sigma(f)\}.\] We denote \(\overline{\mathcal{L}}[f]\) this extension. Compute \(\overline{\mathcal{L}}[f_0]\).

9. Compute for any \(t\geq 0\) and any \(r> \sigma(f)\) the line integral \[ \frac{1}{i2\pi}\int_{r[\circlearrowleft]} \overline{\mathcal{L}}[f] e^{st} \, ds. \]

Answers

1. If \(\sigma(f) < 0\) then \(\sigma = 0\) is an exponential bound of \(f\): there is a \(\kappa > 0\) such that \(\forall\, z \in \mathbb{C}\), \(|f(z)| \leq \kappa\). The function \(f\) is entire and bounded, thus by Liouville’s theorem, it is identically zero.

2. For any complex number \(z\), \[ |f_0(z)| = |e^{-z}| = e^{-\mathrm{Re}(z)} \leq e^{|z|} \] hence \(\sigma = 1\) is an exponential bound of \(f_0\). Conversely, if \(|f_0(z)| \leq \kappa e^{\sigma|z|}\) holds for any complex number \(z\), then \[ \forall \, x \geq 0, \; f_0(-x)=e^{x} \leq \kappa e^{\sigma |-x|} = \kappa e^{\sigma x}, \] thus \(\sigma \geq 1\). Finally, \(\sigma(f_0) = 1\).

3. If for some \(\kappa > 0\) and \(\sigma \geq 0\), \(\forall \, n \in \mathbb{N}\), \(|n! a_n| \leq \kappa \sigma^n\), then \[ |f(z)| \leq \kappa \sum_{n=0}^{+\infty} \frac{(\sigma |z|)^n}{n!} = \kappa e^{\sigma |z|} \] hence \(f\) is of exponential type and \(\sigma(f) \leq \sigma\).

4. For any \(n \in \mathbb{N}\) and \(r>0\), \[ a_n = \frac{1}{i2\pi} \int_{r[\circlearrowleft]}\frac{f(z)}{z^{n+1}}, \] hence if \(\forall \, z \in \mathbb{C}\), \(|f(z)| \leq \kappa e^{\sigma |z|}\), by the M-L inequality, \[ |a_n| \leq \frac{1}{2\pi} \kappa \frac{e^{\sigma r}}{r^{n+1}}\times 2\pi r = \kappa \frac{e^{\sigma r}}{r^n}. \] If \(n=0\), the infimum of the right-hand side with respect to \(r>0\) is \(\kappa\). Otherwise, the right-hand side is a differentiable function of \(r>0\) that tends to \(+\infty\) when \(r\to 0\) or \(r\to +\infty\). Therefore, it has a minimum at a \(r>0\) that satisfies \[ \frac{d}{dr} \ln \left(\kappa \frac{e^{\sigma r}}{r^n}\right) = 0. \] Given that \[ \frac{d}{dr} \ln \left(\kappa \frac{e^{\sigma r}}{r^n}\right) = \frac{d}{dr} \left(\ln \kappa + \sigma r - n \ln r\right) =\sigma - \frac{n}{r}, \] we have \[ |a_n| \leq \min_{r>0} \kappa \frac{e^{\sigma r}}{r^n} = \kappa \left(\frac{e \sigma}{n}\right)^n. \]

5. Let \(\sigma > \sigma(f)\). Let \(\lambda < 1\) such that \(\lambda \sigma > \sigma(f)\); there is a \(\kappa > 0\) such that \[ \forall \, z \in \mathbb{C}, \; |f(z)| \leq \kappa e^{\lambda\sigma|z|}. \] The result of the previous question yields for any \(n>0\) \[ |a_n| \leq \kappa \left(\frac{e \lambda \sigma}{n}\right)^n \] and \(|a_0| \leq \kappa\), therefore \[ \forall \, n \in \mathbb{N}^*, \; |n ! a_n| \leq \kappa \left[ {\sqrt{2\pi n}} \lambda^n(1+\epsilon_n) \right] \sigma^n \; \; \mathrm{ with } \lim_{n \to +\infty} \epsilon_n = 1. \] The sequence \[ \left|\kappa \left[ {\sqrt{2\pi n}} \lambda^n(1+\epsilon_n) \right] \right|, \; n \in \mathbb{N}^* \] converges to \(0\) when \(n\) tends to \(+\infty\) and thus has some upper bound \(\kappa'\) that we may select greater than \(\kappa\). Finally, \(|n! a_n| \leq \kappa' \sigma^n\) for any \(n\in \mathbb{N}\).

6. The sum \(\sum_{n=0}^{+\infty} |a_n| z^n\) is convergent for any \(z \in \mathbb{C}\) because the Taylor expansion of \(f\) is absolutely convergent on the whole complex plane; the function \[ g: z \in \mathbb{C} \mapsto \sum_{n=0}^{+\infty} |a_n| z^n \] is actually an entire function.

   For any \(\sigma\) such that \(\sigma(f) < \sigma\), by question 5, there is a \(\kappa > 0\) such that \(|n! a_n| \leq \kappa \sigma^n\) and thus \(|n! |a_n|| \leq \kappa \sigma^n\). By question 3, the function \(g\) is of exponential type and \(\sigma(g) \leq \sigma\); thus \(\sigma(g) \leq \sigma(f)\). The converse inequality may be proved by the same kind of argument (or directly as \(|f(z)| \leq g(|z|)\)).

7. For any \(\sigma \in \mathbb{R}\), \(e^{-\sigma t} f_0(t) = e^{-(\sigma + 1)t},\) hence the function \(t \in \mathbb{R}_+ \mapsto e^{-\sigma t} f_0(t)\) is integrable if and only if \(\sigma > -1\). Therefore, the Laplace transform of the function \(f_0\) is defined exactly on \(\{s \in \mathbb{C} \; | \; \mathrm{Re} \, \, s > -1\}.\) For any such \(s\), we have \[ \mathcal{L}[f_0](s) = \int_{0}^{+\infty} e^{-t} e^{-st} \, dt = \left[-\frac{e^{-(s+1)t}}{s+1}\right]^{+\infty}_0 = \frac{1}{s+1}. \] In general, if \(f\) is an entire function of exponential type, then for any complex number \(s\) such that \(\sigma = \mathrm{Re}(s) > \sigma(f)\), there is a \(\epsilon > 0\) such that \(\sigma - \epsilon > \sigma(f)\), hence there is some \(\kappa >0\) such that \[ \forall \,t > 0, \; |f(t) e^{-s t}| = |f(t)|e^{-\sigma t} \leq \kappa e^{-\epsilon t}, \] thus the function \(t \in \mathbb{R}_+ \mapsto f(t) e^{-st}\) is integrable.

8. If \(\mathrm{Re}(s) > \sigma(f)\), we have \[ \mathcal{L}[f](s) = \int_{\mathbb{R}_+} \left[ \sum_{n=0}^{+\infty} a_n t^n \right] e^{-st} \, dt. \] For any \(m \in \mathbb{N}\), \[ \left| \left[ \sum_{n=0}^{m} a_n t^n \right] e^{-st} \right| \leq \left[ \sum_{n=0}^{+\infty} |a_n| t^n \right] e^{-(\mathrm{Re}(s)) t} = g(t) e^{-(\mathrm{Re}(s)) t}. \] Therefore, as \(\sigma(g) = \sigma(f)\), for any \(\sigma>0\) such that \(\sigma(f) < \sigma < \mathrm{Re}(s)\), there is a \(\kappa > 0\) such that \[ \left| \left[ \sum_{n=0}^{m} |a_n| t^n \right] e^{-st} \right| \leq \kappa e^{\sigma t} e^{-(\mathrm{Re}(s))t} = \kappa e^{-(\mathrm{Re}(s) - \sigma) t}. \] The right-hand side of this inequality is an integrable function of \(t\); by the dominated convergence theorem \[ \mathcal{L}[f](s) = \int_{\mathbb{R}_+} \left[ \sum_{n=0}^{+\infty} a_n t^n \right] e^{-st} \, dt = \sum_{n=0}^{+\infty} a_n \left[ \int_{\mathbb{R}_+} t^n e^{-st} \, dt \right]. \] For any natural number \(n\), we have \[ \int_{\mathbb{R}_+} t^n e^{-st} \, dt = \mathcal{L}[t\mapsto t^n](s) = \frac{n!}{s^{n+1}} \] and therefore \[ \mathcal{L}[f](s) = \sum_{n=0}^{+\infty} \frac{n! a_n}{s^{n+1}}. \]

   Now, the right-hand side of this equation is a Laurent series centered on the origin which is convergent whenever \(\mathrm{Re}(s) > \sigma(f)\). Consequently, the sum \(g(s)\) is actually convergent and holomorphic in the annulus \(A(0, \sigma(f), +\infty)\).

   Now, the set \[ \Omega = \mathrm{dom} \, \mathcal{L}[f] \cup A(0,\sigma(f), +\infty). \] is open and connected as the union of two connected sets with a non-empty intersection. We may define on \(\Omega\) the function \[ \overline{\mathcal{L}}[f](s) = \left| \begin{array}{cl} \mathcal{L}[f](s) & \mbox{ if } \, s\in \mathrm{dom} \, \mathcal{L}[f], \\ g(s) & \mbox{ if } |s| > \sigma(f). \end{array} \right. \] This definition is non-ambiguous: since \(\mathcal{L}[f](s)\) and \(g\) are holomorphic and identical on the non-empty set \(\{s \in \mathbb{C} \; | \; \mathrm{Re} \, s > \sigma(f)\}\), by the isolated zeros theorem, they have the same values on the connected open set \[\mathrm{dom} \, \mathcal{L}[f] \cap A(0,\sigma(f), +\infty).\]

   The function \(\overline{\mathcal{L}}[f]\) clearly is a holomorphic extension of \(\mathcal{L}[f]\) to \(\Omega\). This extension is unique: the set \(\Omega\) is connected as the union of two connected sets with a non-empty intersection, therefore the isolated zeros theorem also applies.

   For \(f=f_0\), we have \[ \mathrm{dom} \, \mathcal{L}[f] = \{ s \in \mathbb{C} \; | \; \mathrm{Re}(s) > -1 \} \] and as \(\sigma(f_0) = 1\), \[ A(0, \sigma(f_0), +\infty) = \{s \in \mathbb{C} \; | \; |s| > 1\}, \] thus \(\Omega = \mathbb{C} \setminus \{-1\}\). The function \[ s \in \mathbb{C} \setminus \{-1\} \mapsto \frac{1}{s+1} \] is holomorphic. As it extends the original Laplace transform of \(f_0\), this is the extended Laplace transform of \(f_0\).

9. For any \(t\geq 0\) and any \(r> \sigma(f)\), \[ \frac{1}{i2\pi}\int_{r[\circlearrowleft]} \overline{\mathcal{L}}[f] e^{st} \, ds = \frac{1}{i2\pi}\int_{r[\circlearrowleft]} \left[ \sum_{n=0}^{+\infty} \frac{n! a_n}{s^{n+1}} \right] e^{st} \, ds. \] The series \(\sum_{n=0}^{+\infty} {n! a_n}/{s^{n+1}}\) is convergent, uniformly with respect to \(s\) on the circle of radius \(r\) centered on the origin, hence \[ \frac{1}{i2\pi}\int_{r[\circlearrowleft]} \overline{\mathcal{L}}[f] e^{st} \, ds = \sum_{n=0}^{+\infty} n! a_n \left[ \frac{1}{i2\pi}\int_{r[\circlearrowleft]} \frac{e^{st}}{s^{n+1}} \, ds \right]. \] The function \(s \in \mathbb{C}^* \mapsto {e^{st}}/{s^{n+1}}\) is complex-differentiable, therefore \[ \frac{1}{i2\pi}\int_{r[\circlearrowleft]} \frac{e^{st}}{s^{n+1}} \, ds = \mathrm{res} \left(s \mapsto \frac{e^{st}}{s^{n+1}}, 0\right). \] Since \[ \lim_{s \to 0} s^{n+1} \frac{e^{st}}{s^{n+1}} = 1, \] the origin is a pole of order \(n+1\) of this function and \[ \mathrm{res} \left(s \mapsto \frac{e^{st}}{s^{n+1}}, 0\right) = \lim_{s \to 0} \frac{1}{n!}\frac{d^n}{ds^n}e^{st} = \frac{t^n}{n!}. \] Finally, \[ \frac{1}{i2\pi}\int_{r[\circlearrowleft]} \overline{\mathcal{L}}[f] e^{st} \, ds = \sum_{n=0}^{+\infty} n! a_n \frac{t^n}{n!} = \sum_{n=0}^{+\infty} a_n t^n = f(t). \]