Definite Integral to Infinity of Reciprocal of 1 plus Power of x

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Theorem

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

where:

$n$ is a real number greater than 1
$\csc$ is the cosecant function.


Corollary

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


Proof 1

From Euler's Reflection Formula:

$\map \Gamma {\dfrac 1 n} \map \Gamma {1 - \dfrac 1 n} = \pi \map \csc {\dfrac \pi n}$

Then:

\(\ds \map \Gamma {\frac 1 n} \map \Gamma {1 - \frac 1 n}\) \(=\) \(\ds \frac {\map \Gamma {\frac 1 n} \map \Gamma {1 - \frac 1 n} } {\map \Gamma {1 - \frac 1 n + \frac 1 n} }\) $\map \Gamma 1 = 1$
\(\ds \) \(=\) \(\ds \map \Beta {\frac 1 n, 1 - \frac 1 n}\) Definition 3 of Beta Function
\(\ds \) \(=\) \(\ds 2 \int_0^{\pi / 2} \paren {\sin \theta}^{\frac 2 n - 1} \paren {\cos \theta}^{1 - \frac 2 n} \rd \theta\) Definition 2 of Beta Function
\(\ds \) \(=\) \(\ds 2 \int_0^{\pi / 2} \paren {\frac {\sin \theta} {\cos \theta} }^{\frac 2 n - 1} \rd \theta\)
\(\ds \) \(=\) \(\ds 2 \int_0^{\pi / 2} \paren {\tan \theta}^{\frac 2 n - 1} \rd \theta\) Definition of Tangent Function

Note we have:

\(\ds \frac {\map \d {\paren {\tan \theta}^{\frac 2 n} } } {\d \theta}\) \(=\) \(\ds \frac {\map \d {\tan \theta} } {\d \theta} \cdot \frac {\map \d {\paren {\tan \theta}^{\frac 2 n} } } {\map \d {\tan \theta} }\) Chain Rule for Derivatives
\(\ds \) \(=\) \(\ds \sec^2 \theta \cdot \frac 2 n \paren {\tan \theta}^{\frac 2 n - 1}\) Derivative of Tangent Function, Derivative of Power
\(\ds \) \(=\) \(\ds \frac 2 n \paren {1 + \tan^2 \theta} \paren {\tan \theta}^{\frac 2 n - 1}\) Difference of Squares of Secant and Tangent
\(\ds \) \(=\) \(\ds \frac 2 n \paren {1 + \paren {\tan^{\frac 2 n} \theta}^n} \cdot \paren {\tan \theta}^{\frac 2 n - 1}\)

As $\theta \nearrow \dfrac \pi 2$, $\tan \theta \to \infty$ and $\tan 0 = 0$, so making a substitution of $x = \paren {\tan \theta}^{\frac 2 n}$ to our original integral:


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\(\ds 2 \int_0^{\pi / 2} \paren {\tan \theta}^{\frac 2 n - 1} \rd \theta\) \(=\) \(\ds 2 \int_0^\infty \paren {\tan \theta}^{\frac 2 n - 1} \frac {\rd x} {\frac 2 n \paren {1 + \paren {\tan^{\frac 2 n} \theta}^n} \cdot \paren {\tan \theta}^{\frac 2 n - 1} }\) Integration by Substitution
\(\ds \) \(=\) \(\ds \frac {2 n} 2 \int_0^\infty \frac {\paren {\tan\theta}^{\frac 2 n - 1} } {\paren {\tan \theta}^{\frac 2 n - 1} } \frac {\rd x} {1 + x^n}\)
\(\ds \) \(=\) \(\ds n \int_0^\infty \frac 1 {1 + x^n} \rd x\)

So we have:

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

Hence:

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

$\blacksquare$


Proof 2

Let $R > 1$ be a real number.

Let:

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


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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.


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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

$\blacksquare$

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