Hint: assume instead that some complex number \(w\) is not in the closure of the image of \(f\); study the function \(z \mapsto 1/(f(z) - w)\) in a neighbourhood of \(a\).

Note that either way, there is a non-negative integer \(p\) and a holomorphic function \(h :\Omega \cup \{a\} \to \mathbb{C}\) such that \(h(a) \neq 0\) and \[ \forall \, z \in \Omega \cup \{a\}, \; g(z) = h(z) (z-a)^p. \] As the function \(g\) has no zero on \(\Omega\), the function \(h\) has no zero on \(\Omega \cup \{a\}\); the function \(1/h\) is defined and holomorphic on \(\Omega \cup \{a\}\), \(1/h(a) \neq 0\) and \[ \forall \, z \in \Omega, \; f(z) = w + \frac{1}{g(z)} = w + \frac{1}{h(z)} \frac{1}{(z-a)^p}. \] Therefore, the point \(a\) is either a removable singularity of \(f\) (if \(p=0\)), or a pole of order \(p\) (if \(p\geq 1)\).

Suppose that there is a positive integer \(p\) such that \[ f(z) = f(a) + g(z) (z-a)^p \] for some holomorphic function \(g:\Omega \to \mathbb{C}\) such that \(g(a) \neq 0\); there is a function \(\epsilon_a : \Omega \to \mathbb{C}\) such that \(\epsilon_a(z) \to 0\) when \(z \to a\) and \[ f(z) = f(a) + g(a) (z-a)^p + \epsilon_a(z)(z-a)^p \]

Assume that \(f(a) \neq 0\) (if \(f(a)=0\), it is plain that \(|f(a)|=0\) cannot be a local maximum of \(|f|\) at \(a\)). Let \(\alpha\), \(\beta\) and \(\gamma\) be some real numbers such that \[ f(a) = |f(a)| e^{i\alpha}, \; g(a) = |g(a)|e^{i\beta}, \; \gamma = \frac{\theta - \alpha}{p}. \] For small enough values \(r > 0\), we have \[ |f(a + r e^{i \gamma}) - (|f(a)| + |g(a)|r^p) e^{i\alpha}| \leq |\epsilon_a(a + r e^{i\gamma})| r^p \leq \frac{|g(a)|}{2} r^p, \] which yields \[ |f(a + r e^{i \gamma})| \geq |f(a)| + |g(a)| r^p - \frac{|g(a)|}{2} r^p > |f(a)|. \] Therefore \(f\) has no maximum at \(a\).

Questions

1. Find the domain in the complex plane of the function \[ \Pi: z \mapsto \int_0^{+\infty} t^z e^{-t} \, dt \] and show that it is holomorphic.

2. Prove that whenever \(\Pi(z)\) is defined, \(\Pi(z+1)\) is also defined and \[ \Pi(z+1) = (z+1)\Pi(z). \] Compute \(\Pi(n)\) for every \(n \in \mathbb{N}.\)

3. Let \(\Omega\) be an open connected subset of the complex plane that contains the domain of \(\Pi\) and such that \(\Omega + 1 \subset \Omega\). Prove that if \(\Pi\) has a holomorphic extension on \(\Omega\) (still denoted \(\Pi\)), it is unique and satisfies the functional equation \[ \forall \, z \in \Omega, \; \Pi(z+1) = (z+1)\Pi(z). \]

4. Prove the existence of such an extension \(\Pi\) on \[ \Omega = \mathbb{C} \setminus \{k \in \mathbb{Z} \; | \; k < 0 \}. \]

5. Show that every negative integer is a simple pole of \(\Pi\); compute the associated residue.

Answers

1. The function \(t \in \mathbb{R}_+^* \mapsto t^z e^{-t}\) is continuous and thus measurable. Additionally, for any \(t > 0\), \[ |t^z e^{-t}| = |e^{z \ln t} e^{-t}| = e^{(\mathrm{Re} \,z) \ln t}e^{-t} = t^{\mathrm{Re} \,z} e^{-t}, \] hence it is integrable if and only if \(\mathrm{Re} \,z > -1\): the domain of \(\Pi\) is \[ \{z \in \mathbb{C} \; | \; \mathrm{Re} \,z > -1\} \] and it is open. Now, let \(z\) and \(h\) be complex numbers in this domain; the associated difference quotient satisfies \[ \begin{split} \frac{\Pi(z+h) - \Pi(z)}{h} &= \int_0^{+\infty} \frac{t^{z+h} - t^z}{h} e^{-t} dt \\ &= \int_0^{+\infty} \frac{t^h - 1}{h} t^z e^{-t} dt \\ &= \int_0^{+\infty} \frac{e^{h \ln t} - 1}{h} t^z e^{-t} dt \\ &= \int_0^{+\infty} \left[\frac{e^{h \ln t} - 1}{h \ln t}\right] t^z \ln t \, e^{-t} dt \end{split} \] The integrand converges pointwise when \(h\to 0\): \[ \forall \, t > 0, \; \lim_{h \to 0} \left[\frac{e^{h \ln t} - 1}{h \ln t}\right] t^z \ln t \, e^{-t} = t^z \ln t \, e^{-t}. \] Additionally, we have \[ \forall \, z \in \mathbb{C}^*, \; \left|\frac{e^z - 1}{z}\right| \leq e^{|z|}; \] indeed, for any nonzero complex number \(z\), the Taylor expansion of \(e^z\) at the origin provides \[ \left|\frac{e^z - 1}{z}\right| = \left|\sum_{n=0}^{+\infty} \frac{1}{(n+1)!} z^n\right| = \sum_{n=0}^{+\infty} \frac{1}{(n+1)!} |z|^n \leq \sum_{n=0}^{+\infty} \frac{1}{n!} |z|^n. \] Hence, \[ \left| \frac{e^{h \ln t} - 1}{h \ln t} \right| \leq e^{|h||\ln t|} \leq \max(t^{|h|}, t^{-|h|}) \] and our integrand is dominated by \[ \max(t^{z+|h|}, t^{z-|h|}) \ln t \, e^{-t} \] which is integrable whenever \(\mathrm{Re} (z - |h|) > -1\). Finally, Lebesgue’s dominated convergence theorem applies and \(\Pi\) is holomorphic.

2. If \(\mathrm{Re} \,z > -1\), then \(\mathrm{Re} (z +1) > -1\) and \[ \Pi(z+1) = \int_0^{+\infty} t^{z+1} e^{-t} \, dt. \] By integration by parts, \[ \begin{split} \Pi(z+1) &= [t^{z+1}(-e^{-t})]^{+\infty}_0 - \int_0^{+\infty} (z+1) t^z (-e^{-t})\, dt \\ &= (z+1) \Pi(z). \end{split} \]

   We have \[ \Pi(0) = \int_0^{+\infty} e^{-t} \, dt = [-e^{-t}]_0^{+\infty} = 1 \] and hence, by induction, \(\Pi(n) = n!\) for any \(n \in \mathbb{N}\).

3. There is at most one holomorphic extension \(\Pi\) of the original function to the connected open set \(\Omega\) by the isolated zeros theorem (two extensions would be identical on the original domain of \(\Pi\), which is a non-empty open set: the set of zeros of their difference would not be isolated).

   It is plain that the function \(z \mapsto \Pi(z+1) - (z+1)\Pi(z)\) is defined and holomorphic on \(\Omega\), a connected open set of the plane. Similarly, by the isolated zeros theorem, it is identically zero and hence the functional equation \(\Pi(z+1) = (z+1)\Pi(z)\) holds on \(\Omega\).

4. We may define the extension \(\Pi(z)\) as \[ \Pi(z) = \frac{\Pi(z+n)}{(z+1)(z+2)\cdots(z+n)} \] for any natural number \(n\) such \(\mathrm{Re} (z+n) > -1\). This definition does not depend on the choice of \(n\): if \(m > n\), we have \(\mathrm{Re} (z+m) > -1\) and \[ \Pi(z+m) = \Pi(z+n) \times (z+n+1) \cdots (z+m) , \] hence \[ \frac{\Pi(z+m)}{(z+1)(z+2)\cdots(z+m)} = \frac{\Pi(z+n)}{(z+1)(z+2)\cdots(z+n)}. \] It is plain that this extension of the original function \(\Pi\) is holomorphic.

5. Let \(n\) be a positive integer. Let \(z\) be a complex number such that \(|z-(-n)| < 1\); it satisfies \(\mathrm{Re}(z + n) > -1\) and thus \[ \Pi(z) = \frac{\Pi(z+n)}{(z+1)(z+2)\cdots(z+n)}. \] Consequently, \[ (z - (-n)) \Pi(z) = \frac{\Pi(z+n)}{(z+1)(z+2)\cdots(z+n-1)} \] and \[ \lim_{z \to -n} (z - (-n)) \Pi(z) = \frac{\Pi(0)}{(-n-1)(-n-2)\cdots(-1)} = \frac{(-1)^{n-1}}{(n-1)!}. \] As this number differ from zero, \(z=-n\) is a simple pole of \(\Pi\) and \[ \mbox{res}(\Pi, -n) = \frac{(-1)^{n-1}}{(n-1)!}. \]

The function \(z \mapsto \frac{\sin \pi z}{\pi z}\) is therefore defined and holomorphic in \(\mathbb{C}^*\) where its Laurent expansion is \[ \frac{\sin \pi z}{\pi z} = \sum_{n=0}^{+\infty} \frac{(-1)^{n}\pi^{2n}}{(2n+1)!}z^{2n}. \] The series on the right-hand side of this equation has no negative power of \(z\): it is a power series that converges for any \(z \in \mathbb{C}^*\), thus its open disk of convergence is actually \(\mathbb{C}\). Its limit is a holomorphic function that extends \(z \mapsto \frac{\sin \pi z}{\pi z}\) to \(\mathbb{C}\), hence \(0\) is a removable singularity of this function (and its residue is \(0\)).

The singularities of \(z \mapsto {1}/{(\sin \pi z)^2}\) are the zeros of \(z \in \mathbb{C} \mapsto \sin \pi z\): the integers. The function is invariant if we substitute \(z + k\) to \(z\) for any \(k \in \mathbb{Z}\), hence we may limit our analysis of the singularities to the origin. If \(z\) is not an integer, we have \[ \frac{1}{(\sin \pi z)^2} = \frac{1}{\pi^2z^2} \left(\frac{\pi z}{\sin \pi z}\right)^2. \] The function \(z \mapsto ({\pi z}/{\sin \pi z})^2\) has a removable singularity at the origin and the value of its holomorphic extension at the origin is nonzero (it is \(1\)), thus the origin is a double pole of the function. We have therefore \[ \mathrm{res} \left(z \mapsto \frac{1}{(\sin \pi z)^2}, 0\right) = \lim_{z \to 0} \left[ \frac{z^2}{2} \frac{1}{(\sin \pi z)^2} \right]'. \] We have \[ \left[ \frac{z^2}{2} \frac{1}{(\sin \pi z)^2} \right]' = \frac{1}{\pi} \left(\frac{(\pi z)\sin \pi z - (\pi z)^2 \cos \pi z}{(\sin \pi z)^3} \right). \] The Taylor expansions of the functions \(\sin\) and \(\cos\) on \(\mathbb{C}\) provide \[ \sin w = w \left(\sum_{n=0}^{+\infty} \frac{(-1)^{n}}{(2n+1)!} w^{2n} \right) = w - \frac{w^3}{6} + w^5 \left(\sum_{n=2}^{+\infty} \frac{(-1)^{n}}{(2n+1)!} w^{2n-4}\right) \] and \[ \cos w = 1 - \frac{w^2}{2} + w^4 \left(\sum_{n=2}^{+\infty} \frac{(-1)^{n}}{(2n)!} w^{2n-4}\right), \] thus there are entire functions \(f\) and \(g\) such that \[ w \sin w - w^2 \cos w = \left(w^2 - \frac{1}{6} w^4\right) - \left(w^2 -\frac{1}{2} w^4\right) + w^6 f(w) \] and \[ (\sin w)^3 = w^3 g(w), \; g(0) = 1. \] Consequently, \[ \mathrm{res} \left(z \mapsto \frac{1}{(\sin \pi z)^2}, 0\right) = \lim_{w \to 0} \frac{1}{\pi} \frac{w/3 + w^3 f(w)}{g(w)} = 0. \]

Alternatively, to compute the residue, we may notice that if \(z\) is not an integer \[ \frac{1}{(\sin \pi z)^2} = \frac{1}{(\sin \pi (-z))^2}, \] thus if \(\sum_{n=-\infty}^{+\infty} a_n z^n\) is the Laurent expansion of the right-hand side in \(D(0,1)\setminus\{0\}\), the Laurent expansion \(\sum_{n=-\infty}^{+\infty} (-1)^n a_n z^n\) is also valid in the same annulus. The uniqueness of the Laurent expansion yields \(a_{n} = 0\) for every odd \(n\), thus the residue of the function at the origin – which is \(a_{-1}\) – is zero.

The function \(z \mapsto \sin \frac{\pi}{z}\) is defined and holomorphic on \(\mathbb{C}^*\). It has a Laurent expansion in this annulus, which is \[ \sin \frac{\pi}{z} = \sum_{n=0}^{+\infty} \frac{(-1)^{n} \pi^{2n+1}}{(2n+1)!} z^{-(2n+1)}. \] There are an infinite number of nonzero coefficients associated with negative powers of \(z\), thus \(0\) is an essential singularity of this function. Its residue at \(0\) is the coefficient of \(z^{-1}\), which is \(\pi\).

The zeros of \(z \in \mathbb{C} \mapsto \sin \pi z\) are the integers, thus \(z \mapsto {1}/{\sin \frac{\pi}{z}}\) is defined and holomorphic on the open set \(\Omega = \mathbb{C}^{*} \setminus \{1/k \; | \; k \in \mathbb{Z}^*\}\). We can write the function as the quotient of \(f(z) = 1\) and \(g(z) = \sin \frac{\pi}{z}\). The functions \(f\) and \(g\) are defined and holomorphic in \(\mathbb{C}^*\) and \[g'(z) = \left(\cos \frac{\pi}{z}\right)\left(-\frac{\pi}{z^2}\right).\] Thus, for any \(k \in \mathbb{Z}^*\), \(1/k\) is a simple pole of \(z \mapsto {1}/{\sin \frac{\pi}{z}}\) and \[ \mathrm{res} \left( z \mapsto \frac{1}{\sin \frac{\pi}{z}}, \frac{1}{k} \right) = \frac{1}{(\cos \frac{\pi}{k^{-1}})(-\frac{\pi}{{(k^{-1}})^2})} = \frac{(-1)^{k+1}}{\pi k^2}. \] The origin \(z=0\) is also singularity of \(z \mapsto {1}/{\sin \frac{\pi}{z}}\), but it is not isolated, thus its residue is not defined.

Questions

1. For any \(n \geq 2\), compute \[ \int_0^{+\infty} \frac{dx}{1 + x^n}. \]

2. Compute \[ \int_0^{+\infty} \frac{\sqrt x}{1+x+x^2} \, dx. \]

Answers

1. Let \(f\) be the function \(z\mapsto {1}/{(1+z^n)}\), defined and holomorphic on \[ \Omega = \mathbb{C} \setminus \left\{ e^{\frac{i(2k+1)\pi}{n}} \; \left| \vphantom{e^{\frac{(2k+1)\pi}{n}}} \; k \in \{0, \dots, n-1\}\right. \right\}. \] Let \(r>1\) and define the rectifiable paths \(\gamma_1\), \(\gamma_2\) and \(\gamma_3\) as \[ \gamma_1 = [0 \to r], \; \gamma_2 = r e^{i[0 \to 2\pi/n]}, \; \gamma_3 = [re^{i2\pi/n} \to 0], \] then set \(\gamma = \gamma_1 \;| \; \gamma_2 \;|\; \gamma_3.\) It is plain that \[ \lim_{r \to 0} \int_{\gamma_1}\frac{dz}{1+z^n} = \int_0^{+\infty} \frac{dx}{1 + x^n}. \] Similarly, \[ \int_{\overleftarrow{\gamma_3}}\frac{dz}{1+z^n} = \int_0^1 \frac{re^{i\frac{2\pi}{n}}dt}{1 + (rt)^n (e^{i\frac{2\pi}{n}})^n} = e^{i\frac{2\pi}{n}} \int_0^r \frac{dx}{1 + x^n}, \] thus \[ \lim_{r \to 0} \int_{\gamma_3}\frac{dz}{1+z^n} = - e^{i\frac{2\pi}{n}} \int_0^{+\infty} \frac{dx}{1 + x^n}. \] Finally, by the M-L inequality, \[ \left| \int_{\gamma_2}\frac{dz}{1+z^n} \right| \leq \frac{1}{r^n - 1} \times \left(\frac{2\pi}{n} r \right), \] hence \[ \lim_{r \to +\infty} \int_{\gamma_2}\frac{dz}{1+z^n} = 0. \] On the other hand, the complex number \(e^{i\frac{\pi}{n}}\) is the unique singularity of \(f\) in the interior of \(\gamma\); more precisely, we have \(\mathrm{ind}(\gamma, e^{i\frac{\pi}{n}})=1\). The function \(f\) is the quotient of the holomorphic functions \(p: z \in \mathbb{C} \mapsto 1\) and \(q: z \in \mathbb{C} \mapsto 1+z^n\); the derivative of \(q\) at this singularity is \[ q'(e^{i\frac{\pi}{n}}) = n (e^{i\frac{\pi}{n}})^{n-1} = n (e^{i\frac{\pi}{n}})^{n} e^{-i\frac{\pi}{n}} = -n e^{-i\frac{\pi}{n}}, \] thus \[ \mathrm{res}(f, e^{i\frac{\pi}{n}}) = \frac{p(e^{i\frac{\pi}{n}})}{q'(e^{i\frac{\pi}{n}})} = -\frac{e^{i\frac{\pi}{n}}}{n} \] Given these results, the residue theorem provides \[ \left(1 - e^{i\frac{2\pi}{n}}\right) \int_0^{+\infty} \frac{dx}{1 + x^n} = (i2\pi) \times \left(-\frac{e^{i\frac{\pi}{n}}}{n}\right) \] or equivalently, \[ \int_0^{+\infty} \frac{dx}{1 + x^n} = \frac{\pi}{n} \frac{2i}{e^{i\frac{\pi}{n}} - e^{-i\frac{\pi}{n}}} = \frac{\frac{\pi}{n}}{\sin \frac{\pi}{n}}. \]

2. Let \(\log_0\) be the function defined on \(\mathbb{C} \setminus \mathbb{R}_+\) by \[\log_0 z = \log (-z) + i\pi.\] This function is an analytic choice of the logarithm on \(\mathbb{C} \setminus \mathbb{R}_+\): it is holomorphic and \(\exp \circ \log_0\) is the identity. It also satisfies \[ \log_0 r e^{i\theta} = (\ln r) + i\theta, \; r >0, \; \theta \in \left]0,2\pi\right[. \] We use this function to define \[ f: z \mapsto \frac{e^{\frac{1}{2}\log_0 z}}{1+z+z^2}. \] The roots of the polynomial \(z\mapsto 1+z+z^2\) are \(j\) and \(j^2\), where \(j = e^{i\frac{2\pi}{3}}\), thus \(f\) is defined and holomorphic in \(\Omega = \mathbb{C} \setminus \mathbb{R}_+ \setminus \{j,j^2\}\).

   Now, let \(r>1\) and \(0 < \alpha < 2\pi/3\); we define four rectifiable paths that depend on \(r\) and \(\alpha\): \[ \begin{split} \gamma_1 &= [r^{-1} e^{i\alpha} \to r e^{i\alpha}], \\ \gamma_2 &= r e^{i[\alpha \to 2\pi-\alpha]},\\ \gamma_3 &= [re^{i(2\pi - \alpha)} \to r^{-1}e^{i(2\pi - \alpha)}],\\ \gamma_4 &= r^{-1} e^{i[2\pi -\alpha \to \alpha]}. \end{split} \] We also consider their concatenation \[ \gamma = \gamma_1 \; | \; \gamma_2 \; | \; \gamma_3 \; | \; \gamma_4. \]

   We have \[ \begin{split} \int_{\gamma_1} f(z) \, dz &= \int_{r^{-1}}^r \frac{e^{\frac{1}{2}((\ln x) + i \alpha)}}{1+x e^{i\alpha}+x^2 e^{i2\alpha}}\, e^{i\alpha}dx\\ &= e^{i3{\alpha}/{2}} \int_{r^{-1}}^r \frac{\sqrt{x}}{1+x e^{i\alpha}+x^2 e^{i2\alpha}}\, dx \end{split} \] and thus by the dominated convergence theorem¹ \[ \lim_{\alpha \to 0} \int_{\gamma_1} f(z) \, dz = \int_{r^{-1}}^{r} \frac{\sqrt{x}}{1+x+x^2} \, dx. \]

   Similarly, \[ \begin{split} \int_{\gamma_3^{\leftarrow}} f(z) \, dz &= \int_{r^{-1}}^r \frac{e^{\frac{1}{2}((\ln x) + i(2\pi-\alpha))}}{1+x e^{-i\alpha}+x^2 e^{-i2\alpha}}\, e^{-i\alpha}dx\\ &= - e^{-i{3\alpha}/{2}} \int_{r^{-1}}^r \frac{\sqrt{x}}{1+x e^{-i\alpha}+x^2 e^{-i2\alpha}}\, dx \end{split} \] and thus by the dominated convergence theorem \[ \lim_{\alpha \to 0} \int_{\gamma_3} f(z) \, dz = \int_{r^{-1}}^{r} \frac{\sqrt{x}}{1+x+x^2} \, dx \] On the other hand, \[ \left|e^{\frac{1}{2} \log_0 z}\right| = e^{\mathrm{Re}(\frac{1}{2} \log_0 z)} = e^{\frac{1}{2} \ln |z|} = {|z|}^{\frac{1}{2}}; \] by the M-L inequality, this equality provides \[ \left| \int_{\gamma_2} f(z) \, dz \right| \leq \frac{{r}^{\frac{1}{2}}}{-1-r+r^2}\times 2(\pi-\alpha) r \] and \[ \left| \int_{\gamma_4} f(z) \, dz \right| \leq \frac{{r}^{-\frac{1}{2}}}{1-r^{-1}-r^{-2}}\times 2(\pi-\alpha) r^{-1}, \] hence \[ \lim_{r \to +\infty} \left(\lim_{\alpha \to 0} \int_{\gamma_2} f(z) \, dz\right) = \lim_{r \to +\infty} \left(\lim_{\alpha \to 0} \int_{\gamma_4} f(z) \, dz\right) = 0. \]

   Now the function \(f\) is the quotient of the two functions \(z \mapsto e^{\frac{1}{2} \log_0 z}\) and \(z \mapsto 1+z+z^2\), defined and holomorphic in a neighbourhood of the singularities \(j\) and \(j^2\). The derivative of \(z \mapsto 1 + z +z^2\) is \(z \mapsto 1+2z\), it is nonzero at \(j\) and \(j^2\). Thus, \[ \mathrm{res}(f, j) = \frac{e^{\frac{1}{2} \log_0 j}} {1 + 2 j} = \frac{e^{i\frac{\pi}{3}}}{i\sqrt{3}} \] and \[ \mathrm{res}(f, j^2) = \frac{e^{\frac{1}{2} \log_0 j^2}} {1 + 2 j^2} = \frac{e^{i\frac{2\pi}{3}}}{-i\sqrt{3}}. \] The winding number of \(\gamma\) around \(j\) and \(j^2\) is \(1\); by the residue theorem, \[ 2 \int_0^{+\infty} \frac{\sqrt x}{1+x+x^2} \, dx = (i2\pi) (\mathrm{res}(f, j) + \mathrm{res}(f, j^2)) \] or equivalently \[ \int_0^{+\infty} \frac{\sqrt x}{1+x+x^2} \, dx = \frac{\pi}{\sqrt{3}} (e^{i\frac{\pi}{3}} -e^{i\frac{2\pi}{3}}) = \frac{\pi}{\sqrt{3}}. \]