Suppose \(n \geq 2\). Use a wedge of angle \(\frac{2 \pi}{n}\) to evaluate the integral \[ I=\int_{0}^{\infty} \frac{1}{1+x^{n}} d x \]
The current Quals .docx writes the OMML limits as \(-\infty\) to \(\infty\) (Word’s default nary limits). That
integral has a pole on the real axis when \(n\) is odd, and the same sentence asks for
a wedge of angle \(2\pi/n\), which
evaluates the integral from \(0\) to
\(\infty\). The exam as transcribed in
2021_Fall.md.source, and the existing solutions, are \(\int_0^\infty\).
Solution: Newer, sketch
By the ML estimate, \(\int_{C_R} f \to 0\).
The residue contribution: note the simple pole at \(\omega_n \da e^{i\pi \over n}\), \[ \Res_{z=\omega_n} f(z) = {1\over n\omega_n^{n-1}} = {\omega \over n\omega^n} = -{\omega_n \over n} .\]
The segment contributions: \(\int_{\gamma_1}f\to I\), and \[ \int_{\gamma_2}f(z) \dz = \int_\infty^0 {1\over 1 + (\zeta_nt)^n} \zeta_n \dt = -\zeta_n I ,\] so the contour contributions sum to \((1-\zeta_n)I\).
Solving: \[ I = -{2\pi i \omega_n\over n(1-\zeta_n)} = -2\pi i {1\over \inverseof{\omega_n} - \omega_n} = -2\pi i {1\over 2i \sin\qty{\pi \over n}} = {\pi \over n} \csc\qty{\pi \over n} .\]
Solution: Older, detailed
Write \(\omega_{n, k} = \exp\qty{(2k+1)i\pi \over n}\) and factor \(z^n+1\) as \[ z^n+1 = \prod_{1\leq k \leq n}(z-\omega_{n, k}) = (z-e^{i\pi \over n})(z-e^{3i\pi \over n}) \cdots (z-e^{(2n-1)i\pi \over n}) .\] Note that only the root \(e^{i\pi\over n}\) lies in the \(2\pi/n\) wedge, so it is the only (simple) pole of \(f(z) \da {1\over 1+z^n}\) in this region. Since the pole is simple, the residue is computed by the simple-pole formula \(\Res_{z=r_0} f = \lim_{z \to r_0} (z - r_0) f(z)\). Write \(r_0 \da e^{i\pi\over n}\), then by L’Hôpital, \[ \Res_{z = r_0} {1\over 1+z^n} &= \lim_{z\to r_0} {z-r_0 \over 1 + z^n} \\ &= \lim_{z\to r_0} {1\over nz^{n-1}} \\ &= {1\over nr_0^{n-1}} \\ &= {1\over n e^{i\pi (n-1) \over n}} \\ &= \inverseof{n} {\exp\qty{-i\pi (n-1)\over n }} .\]
Take a contour \(\Gamma\) comprised of
\(\gamma_1 = [0, R] \subseteq \RR\)
\(\gamma_2 = \ts{Re^{it} \st t\in [0, 2\pi/n]}\)
\(\gamma_3 = \zeta_n [0, R]\)
By the residue theorem \[ 2\pi i \Res_{z=r_0} f(z) = I \da \int_\Gamma f = \qty{\int_{\gamma_1} + \int_{\gamma_2} + \int_{\gamma_3}}f .\]
Taking orientations into account, \[ \int_{\gamma_3} f = -\zeta_n \int_{\gamma_1} f .\]
\[ \int_{\gamma_2}f\convergesto{R\to\infty}0 .\]
so in the limit we have \[ 2\pi i \Res_{z=r_0}f(z) &= \qty{1 - \zeta_n}\int_{\gamma_1}f \\ \implies \int_{\gamma_1} f &= {2\pi i \Res_{z=r_0}f(z) \over 1 - \zeta_n}\\ &= {2\pi i e^{-\pi (n-1) \over n} \over n\qty{1-e^{2\pi i \over n}}} \\ &= {2\pi i \over n} \inverseof{\left[ e^{i\pi} e^{-i\pi \over n}\qty{1 - e^{2\pi i \over n}} \right]} \\ &= {2\pi i \over n} \inverseof{\left[ -1\qty{e^{-i\pi \over n} - e^{\pi i \over n}} \right]}\\ &= {2\pi i \over n} \inverseof{\left[ 2i \sin\qty{\pi\over n} \right]} \\ &= {\pi \over n\sin\qty{\pi \over n}} .\]
(of claim 1)
Parameterize the curves:
\(\gamma_1 \da \ts{t \st t\in [0, R]}, \dz = \dt\)
\(\gamma_3 \da \ts{t\zeta_n \st t\in [0, R]}, \dz = \zeta_n \dt\)
Then, a direct check: \[ \int_{\gamma_3}f(z) \dz &= \int_0^R {1\over 1 + (\zeta_n t)^n}\zeta_n \dt \\ &= \zeta_n\int_0^R {1\over 1 + t^n}\dt \\ &= \zeta_n \int_{\gamma_1}f(z) \dz .\]
(of claim 2)
Parameterize \(\gamma_2 = \ts{Re^{it} \st t\in [0, 2\pi/n]}\) and apply the ML estimate: \[ {1\over 1 + (Re^{it})^n} \leq {1\over R^n - 1} \implies \int_{\gamma_2}f \leq {1\over R^n - 1} \qty{{2\pi R \over n}} = \bigo(R^{n-1})\convergesto{R\to\infty}0 .\]