# Exercises

## The Weierstrass-Casorati Theorem

### Question

Let $$f: \Omega \to \mathbb{C}$$ be a holomorphic function and let $$a \in \mathbb{C}$$ be an essential singularity of $$f$$. Show that the image of $$f$$ is dense in $$\mathbb{C}$$: $\forall \, w \in \mathbb{C}, \; \forall \, \epsilon > 0, \; \exists \, z \in \Omega, \; |f(z) - w| < \epsilon.$

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 $$a$$.

Assume that the image of $$f$$ is not dense in $$\mathbb{C}$$; let then $$w \in \mathbb{C}$$ be such that $\exists \, \epsilon > 0, \; \forall \, z \in \Omega, \; |f(z) - w| \geq \epsilon.$ The function $$z \in \Omega \mapsto 1/(f(z) - w)$$ is defined and holomorphic. As it satisfies $\forall \, z \in \Omega, \; \left| \frac{1}{f(z) - w} \right| \leq \frac{1}{\epsilon},$ it is also bounded. Thus, the point $$a$$ is a removable singularity of the function, that can be extended as a holomorphic function $$g$$ on $$\Omega \cup \{a\}$$: $\forall \, z \in \Omega, \; g(z) = \frac{1}{f(z) - w}$ By construction, $$g$$ has no zero in $$\Omega$$, thus $$a$$ is either not a zero of $$g$$, or a zero of finite multiplicity. Since $\forall \, z \in \Omega, \; f(z) = w + \frac{1}{g(z)}$ in the first case $$f(z) \to w + 1/g(a)$$ when $$z\to a$$ thus $$a$$ is a removable singularity of $$a$$; in the second one, $$|f(z)| \to +\infty$$ when $$z \to a$$ thus $$a$$ is a pole of $$f$$.

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 $$p\geq 1)$$.

## The Maximum Principle

### Question

Let $$\Omega$$ be an open connected subset of the complex plane and let $$f: \Omega \to \mathbb{C}$$ be a holomorphic function. Show that if $$|f|$$ has a local maximum at some $$a \in \Omega$$, then $$f$$ is constant.

For any holomorphic function $$f: \Omega \to \mathbb{C}$$ and $$a \in \Omega$$, the point $$a$$ is a zero of the holomorphic function $$z \mapsto f(z) - f(a)$$. We will prove shortly that if $$a$$ is a zero of finite multiplicity of this function, $$|f|$$ does not have a local maximum at $$a$$. The conclusion of the proof follows by the Isolated Zeros Theorem.

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 $$a$$.

## The $$\Pi$$ Function

We introduce the $$\Pi$$ function, a holomorphic extension of the factorial.

### 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.

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 $\mbox{res}(\Pi, -n) = \frac{(-1)^{n-1}}{(n-1)!}.$

## Singularities and Residues

### Question

Analyze the singularities (location, type, residues) of $z \mapsto \frac{\sin \pi z}{\pi z}, \; z \mapsto \frac{1}{(\sin \pi z)^2}, \; z \mapsto \sin \frac{\pi}{z}, \; z \mapsto \frac{1}{\sin \frac{\pi}{z}}.$

The function $$z \mapsto \sin \pi z$$ is defined and holomorphic in $$\mathbb{C}$$. Its Taylor expansion, valid for any $$z \in \mathbb{C}$$, is $\sin \pi z = \sum_{n=0}^{+\infty} \frac{(-1)^{n}\pi^{2n+1}}{(2n+1)!}z^{2n+1}.$

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.

## Integrals of Functions of a Real Variable

See “Technologie de calcul des intégrales à l’aide de la formule des résidus” (Demailly 2009, chap. III, sec. 4) for a comprehensive analysis of the computation of integrals with the the residue theorem.

### Questions

1. For any $$n \geq 2$$, compute $\int_0^{+\infty} \frac{dx}{1 + x^n}.$

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

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 $\int_0^{+\infty} \frac{\sqrt x}{1+x+x^2} \, dx = \frac{\pi}{\sqrt{3}} (e^{i\frac{\pi}{3}} -e^{i\frac{2\pi}{3}}) = \frac{\pi}{\sqrt{3}}.$

# References

1. Fonctions Holomorphes et Surfaces de Riemann
Jean-Pierre Demailly, 2009.
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# Notes

1. the function $$(\alpha,x)\mapsto \left|{\sqrt{x}}/{(1+x e^{i\alpha}+x^2 e^{i2\alpha})}\right|$$ is defined and continuous in the compact set $$[0,\pi/2] \times [r^{-1},r]$$, thus it has a finite upper bound.