Definite Integral to Infinity of Reciprocal of 1 plus Power of x/Proof 2

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Theorem

$\ds \int_0^\infty \frac 1 {1 + x^n} \rd x = \frac \pi n \map \csc {\frac \pi n}$


Proof

Let $R > 1$ be a real number.

Let:

$\ds C_R = \set {R e^{i \theta} : 0 \le \theta \le \frac {2 \pi} n}$



Let $L_R$ be the straight line segment from $0$ to $R e^{\frac {2 \pi i} n}$.

Let $\Gamma_R = \closedint 0 R \cup C_R \cup L_R$, traversed anticlockwise.



Then:

\(\ds \oint_{\Gamma_R} \frac 1 {1 + z^n} \rd z\) \(=\) \(\ds \int_0^R \frac 1 {1 + x^n} \rd x + \int_{C_R} \frac 1 {1 + z^n} \rd z + \int_{L_R} \frac 1 {1 + z^n} \rd z\)
\(\ds \) \(=\) \(\ds \int_0^R \frac 1 {1 + x^n} \rd x + \int_{C_R} \frac 1 {1 + z^n} \rd z + e^{\frac {2 \pi i} n} \int_R^0 \frac 1 {1 + e^{2 \pi i} x^n} \rd x\) Definition of Complex Contour Integral
\(\ds \) \(=\) \(\ds \paren {1 - e^{\frac {2 \pi i} n} } \int_0^R \frac 1 {1 + x^n} \rd x + \int_{C_R} \frac 1 {1 + z^n} \rd z\)

We can show the second integral to vanish as $R \to \infty$:

\(\ds \cmod {\int_{C_R} \frac 1 {1 + z^n} \rd z}\) \(\le\) \(\ds \frac {2 \pi R} n \max_{0 \le \theta \le \frac {2 \pi} n} \cmod {\frac 1 {1 + R^n e^{i n \theta} } }\) Estimation Lemma for Contour Integrals
\(\ds \) \(=\) \(\ds \frac {2 \pi R} {n \paren {R^n - 1} }\)
\(\ds \) \(\sim\) \(\ds \frac {2 \pi} {n R^{n - 1} }\)
\(\ds \) \(\to\) \(\ds 0\) as $n > 1$

So we have:

$\ds \oint_{\Gamma_\infty} \frac 1 {1 + z^n} \rd z = \paren {1 - e^{\frac {2 \pi i} n} } \int_0^\infty \frac 1 {1 + x^n} \rd x$


We have that the integrand is meromorphic with a single simple pole at $z = e^{\frac \pi n i}$.

This is contained within the contour

Hence Cauchy's Residue Theorem can be applied to evaluate the left hand side:

\(\ds \oint_{\Gamma_\infty} \frac 1 {1 + z^n} \rd z\) \(=\) \(\ds 2 \pi i \Res {\frac 1 {1 + z^n} } {e^{\frac \pi n i} }\)
\(\ds \) \(=\) \(\ds \frac {2 \pi i} {n e^{\frac \pi n \paren {n - 1} i} }\) Residue at Simple Pole
\(\ds \) \(=\) \(\ds \frac {2 \pi i e^{i \frac \pi n} } {n e^{\pi i} }\)
\(\ds \) \(=\) \(\ds -\frac {2 \pi i e^{i \frac \pi n} } n\) Euler's Identity


So:

$\ds \paren {e^{\frac {2 \pi i} n} - 1} \int_0^\infty \frac 1 {1 + x^n} \rd x = \frac {2 \pi i e^{i \frac \pi n} } n$

giving:

\(\ds \int_0^\infty \frac 1 {1 + x^n} \rd x\) \(=\) \(\ds \frac \pi n \cdot 2 i \cdot \frac {e^{i \frac \pi n} } {e^{\frac {2 \pi i} n} - 1}\)
\(\ds \) \(=\) \(\ds \frac \pi n \cdot \frac {2 i e^{i \frac \pi n} } {e^{\frac \pi n i} \paren {e^{\frac \pi n i} - e^{-\frac \pi n i} } }\)
\(\ds \) \(=\) \(\ds \frac \pi n \cdot \frac {2 i} {e^{\frac \pi n i} - e^{-\frac \pi n i} }\)
\(\ds \) \(=\) \(\ds \frac \pi n \map \csc {\frac \pi n}\) Euler's Cosecant Identity

$\blacksquare$