Application of Cauchy Residue Theorem to evaluation of definitive integral

It ain’t about how hard you can hit. Its about how hard you can get hit, and how much you can take, and keep moving forward, Rocky Balboa

Recall

Laurent’s theorem. Let A = {$z \in \mathbb{C}: R \lt |z -a| \lt S$} be an annular region ($0 \le R \lt S \le \infty$) and let f be analytic on A. Then, $f(z) = \sum_{n=-\infty}^\infty c_n(z -a)^n, \forall z \in A$ where $c_n = \frac{1}{2\pi i}\int_{\gamma} \frac{f(w)}{(w-a)^{n+1}}dw$ and $\gamma$ is a positively oriented circle of radius r centered at a and R < r < S ($\gamma$ is inside the annular region).

Corollary (aka Rosetta Stone of Complex Analysis). Let f be analytic in $\mathbb{B'}(a; r)$. (i) f has a removable singularity at a if and only if in the Laurent's series expansion of f around a ($\sum_{n = -\infty}^\infty (z-a)^n$) all negative coefficients are zero $c_n = 0, \forall n \lt 0$. (ii) f has a pole of order m ($m \ge 1$) at a if and only if the series stop at the negative power -m, $c_{-m} \ne 0, c_n = 0, \forall n \lt -m$; (iii) f has an isolated essential singularity at a if and only if the series has infinitely many negative terms (it never stops), i.e., there is no m such that $c_n = 0, \forall n \lt -m$

The Argument Principle. Suppose f is meromorphic on and inside a simple closed curve $\gamma$ with a finite number of zeroes $a_j, 1 \le j \le l_1$ and poles $b_k, 1 \le k \le l_2$. Suppose none of those zeros and poles lie on $\gamma$. Then, $\frac{1}{2\pi i}\int_{\gamma}\frac{f'(z)}{f(z)}dz = M - N$ where M is the sum of order of zeroes at $a_j, 1 \le j \le l_1$ and N is the sum of order of poles at $b_k, 1 \le k \le l_2$

Application of Cauchy Residue Theorem to evaluation of definitive integral

We want to calculate the improper real integral: $\int_{0}^{\infty} \frac{dx}{1+x^{2n}}, \forall n \in \mathbb{N}, n \gt 1$. We cannot compute this directly using standard Calculus.

Inspired by this integral, we extend the integrand to the complex plane $\frac{1}{1+z^{2n}}$.

We choose a contour $\Gamma$ that consists of three paths (a “pizza slice”, Figure ii):

• 1, The Real Axis: From 0 to R (positive real axis). This is exactly the integral we want (I).
• 2, The Crust: A circular arc $C_R$ from angle $0$ to angle $\frac{\pi}{n}$.
• 3, The Return Trip: A straight line $\gamma$ from the edge back to the origin.

Why this specific angle $\pi/n$? Because the function depends on $z^{2n}$. If we rotate $z$ by an angle of $\pi/n$, then $z^{2n}$ rotates by $2n(\pi/n) = 2\pi$, which brings us back to where we started. This symmetry is absolutely essential for the algebra later.

We want to integrate $\frac{1}{1+z^{2n}}$ around our “Pizza Slice” shape (sector) in the complex plane. We need to find the singularities inside our pizza slice. Singularities occur when the denominator is zero: $1 + z^{2n} = 0 \implies z^{2n} = -1$. In polar form, $-1 = e^{i\pi}, e^{3i\pi}, e^{5i\pi}, \dots$. In polar form, all representations of -1 are $-1=e^{i(\pi +2\pi m)}, m\in \mathbb{Z}$. So, we solve $z^{2n}=e^{i(\pi +2\pi m)}$, the 2n distinct roots are: $z_k = e^{i\frac{(2k+1)\pi}{2n}}, k = 0, 1, 2, \cdots, 2n-1$.

Which one is inside our “pizza” 🍕 slice? Our slice goes from angle $0$ to $\frac{\pi}{n}$ (or $\frac{2\pi}{2n}$).

• The first root (k = 0) has angle $\frac{\pi}{2n}$. Since $\frac{\pi}{2n}$ is exactly halfway between $0$ and $\frac{\pi}{n}$, this pole is inside.
• The next root (k = 1) has angle $\frac{3\pi}{2n}$, which is outside our slice.
• There is exactly one simple pole inside our contour at $a = e^{i\frac{\pi}{2n}}$.

By Cauchy Residue Theorem, $\int_{\Gamma} \frac{1}{1+z^{2n} }dz= 2\pi i \cdot Res\{ \frac{1}{1+z^{2n}}; e^{\frac{i\pi}{2n}} \}$

$\lim_{z \to e^{\frac{i\pi}{2n}}} (z - e^{\frac{i\pi}{2n}}) \frac{1}{1+z^{2n}} =[\text{For simplicity, let's name }a = e^{\frac{i\pi}{2n}}] \lim_{z \to a} \frac{1}{\frac{1+z^{2n}-0}{z-a}}$

Recall the concept of derivative $g'(a) = \lim_{z \to a}\frac{g(z)-g(a)}{z-a}$ where $g(z) = 1 + z^{2n}, g (a) = 0$

We need the residue at $a = e^{i\frac{\pi}{2n}}$.Since it is a simple pole, we use the derivative formula: $\text{Res}(f, a) = \frac{1}{g'(a)} \quad \text{where } g(z) = 1+z^{2n}$

$lim_{z \to a} \frac{1}{\frac{1+z^{2n}-0}{z-a}} = \frac{1}{(1+z^{2n})'|_{e^{\frac{i\pi}{2n}}}} = \frac{1}{2nz^{2n-1}|_{e^{\frac{i\pi}{2n}}}} = \frac{1}{2n(e^{i\pi}e^{\frac{-i\pi}{2n}})} = \frac{e^{\frac{i\pi}{2n}}}{-2n} =\frac{-e^{\frac{i\pi}{2n}}}{2n}$

$2\pi iRes\{ \frac{1}{1+z^{2n}}; e^{\frac{i\pi}{2n}} \} = 2\pi i( \frac{-e^{\frac{i\pi}{2n}}}{2n}) = \frac{-i\pi e^{\frac{i\pi}{2n}}}{n}$

Now we break the loop into its three parts: $\int_{\Gamma} \frac{1}{1+z^{2n}}dz = \int_{0}^{R}\frac{1}{1+x^{2n}}dx + \int_{C_R}\frac{1}{1+z^{2n}}dz + \int_{\gamma}\frac{1}{1+z^{2n}}dz$.

1. The Real Axis. As $R \to \infty$, this becomes our target integral: $\int_0^\infty \frac{dx}{1+x^{2n}} = I$.
2. The Arc or Crust (Path 2). On the outer rim, $|z| = R$. $|1 + z^{2n}| \ge |z|^{2n}-1 = R^{2n}-1$ for large |z| (|z| > 1). $\left| \frac{1}{1+z^{2n}}\right| \leq \frac{1}{R^{2n}-1}\leq \frac{1}{R^{2n}}, |\int_{C_R} \frac{1}{1+z^{2n}}dz| \le \frac{1}{R^{2n}} \int_{C_R}|dz| =[\text{Length of arc = } R \cdot \frac{\pi}{n}] \frac{1}{R^{2n}}\cdot \frac{R\pi}{n} = \frac{\pi}{nR^{2n-1}}$ Since $2n > 1$, as $R \to \infty$ this integral goes to 0.
3. The Return Trip (Path 3). We are integrating along the line $z = r e^{i\frac{\pi}{n}}$ from $R$ down to $0$. In other words, on $\gamma, z = re^{\frac{i\pi}{n}} \text{ where } 0 \le r \le R$

$z = r e^{i\frac{\pi}{n}} \implies dz = e^{i\frac{\pi}{n}} dr$. Then, $1 + z^{2n} = 1 + (r e^{i\frac{\pi}{n}})^{2n} = 1 + r^{2n} e^{i 2\pi} = 1 + r^{2n}$ (The avid reader should not fail to notice how the complex part has completely vanished! This is why we chose this angle and “pizza slice”).

$\int_{\gamma} \frac{dz}{1+z^{2n}} = \int_{R}^{0} \frac{e^{\frac{i\pi}{n}}dr}{1+ (re^{\frac{i\pi}{n}})^{2n}} = -e^{\frac{i\pi}{n}}\int_{0}^{R} \frac{dr}{1+r^{2n}}$ Letting $R \to \infty$, this becomes $-e^{\frac{i\pi}{n}}\int_{0}^{\infty} \frac{dx}{1+x^{2n}} = - e^{i\frac{\pi}{n}} \cdot I$.

L.H.S. = $(1 -e^{\frac{i\pi}{n}})I = (1 -e^{\frac{i\pi}{n}})\int_{0}^{\infty} \frac{dx}{1+x^{2n}}$

R.H.S. = $\frac{-i\pi e^{\frac{i\pi}{2n}}}{n}$

Combining both results: $I = \frac{-\frac{i \pi}{n} e^{i\frac{\pi}{2n}}}{1 - e^{i\frac{\pi}{n}}}$

Multiply the numerator and denominator by $e^{-i\frac{\pi}{2n}}$: $I = \frac{-\frac{i \pi}{n} e^{i\frac{\pi}{2n}} \cdot e^{-i\frac{\pi}{2n}}}{(1 - e^{i\frac{\pi}{n}}) \cdot e^{-i\frac{\pi}{2n}}} = \frac{-\frac{i \pi}{n}}{e^{-i\frac{\pi}{2n}} - e^{i\frac{\pi}{2n}}}$

Swap the sign in the denominator to make it look like sine definition ($e^{ix} - e^{-ix}$), recall $\sin(\theta) = \frac{e^{i\theta} - e^{-i\theta}}{2i} \implies e^{i\theta} - e^{-i\theta} = 2i \sin(\theta)$: $I = \frac{-\frac{i \pi}{n}}{-(e^{i\frac{\pi}{2n}} - e^{-i\frac{\pi}{2n}})} = \frac{\frac{i \pi}{n}}{e^{i\frac{\pi}{2n}} - e^{-i\frac{\pi}{2n}}}= \frac{i \pi / n}{2i \sin(\frac{\pi}{2n})} = \frac{\pi}{2n \sin(\frac{\pi}{2n})}$

$\int_{0}^{\infty} \frac{dx}{1+x^{2n}} = \boxed{\frac{\pi}{2n \sin(\frac{\pi}{2n})}}$