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

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Theorem

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

where:

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


Corollary

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


Proof 1

From Euler's Reflection Formula:

$\displaystyle \Gamma \left({\frac 1 n}\right) \Gamma \left({1 - \frac 1 n}\right) = \pi \csc \left({\frac \pi n}\right)$

Then:

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

Note we have:

\(\displaystyle \frac {\rd \left({\left({\tan \theta}\right)^{\frac 2 n} }\right)} {\rd \theta}\) \(=\) \(\displaystyle \frac {\rd \left({\tan \theta}\right)} {\rd \theta} \cdot \frac {\rd \left({\left({\tan \theta}\right)^{\frac 2 n} }\right)} {\rd \left({\tan \theta}\right)}\) $\quad$ Chain Rule $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \sec^2\theta \cdot \frac 2 n \left({\tan \theta}\right)^{\frac 2 n - 1}\) $\quad$ Derivative of Tangent Function, Derivative of Power $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \frac 2 n \left({1+\tan^2 \theta}\right) \left(\tan \theta\right)^{\frac 2 n - 1}\) $\quad$ Sum of Squares of Sine and Cosine: Corollary 1 $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \frac 2 n \left({1+\left({\tan^{\frac 2 n} \theta}\right)^n}\right) \cdot \left({\tan \theta}\right)^{\frac 2 n - 1}\) $\quad$ $\quad$

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



\(\displaystyle 2 \int_0^{\pi/2} \left({\tan \theta}\right)^{\frac 2 n - 1} \rd \theta\) \(=\) \(\displaystyle 2 \int_0^\infty \left({\tan \theta}\right)^{\frac 2 n - 1} \frac {\rd x} {\frac 2 n \left(1+\left(\tan^{\frac 2 n}\theta\right)^n\right) \cdot \left({\tan \theta}\right)^{\frac 2 n - 1} }\) $\quad$ Integration by Substitution $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \frac {2n} 2 \int_0^\infty \frac{\left({\tan\theta}\right)^{\frac 2 n - 1} } {\left(\tan\theta\right)^{\frac 2 n - 1} } \frac {\rd x} {1 + x^n}\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle n \int_0^\infty \frac 1 {1 + x^n} \rd x\) $\quad$ $\quad$

So we have:

$\displaystyle \pi \csc \left({\frac \pi n}\right) = n \int_0^\infty \frac 1 {1 + x^n} \rd x$

Hence:

$\displaystyle \int_0^\infty \frac 1 {1 + x^n} \rd x = \frac \pi n \csc \left({\frac \pi n}\right)$

$\blacksquare$


Proof 2

Let $R > 1$ be a real number.

Let:

$\displaystyle C_R = \set {Re^{i \theta} : 0 \le \theta \le \frac {2 \pi} n}$

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 anti-clockwise.



Then:

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

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

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

So we have:

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

As the integrand is meromorphic with a single simple pole at $z = e^{\frac \pi n i}$ contained within the contour, the Residue Theorem can be applied to evaluate the LHS:

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

So:

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

Giving:

\(\displaystyle \int_0^\infty \frac 1 {1 + x^n} \rd x\) \(=\) \(\displaystyle \frac \pi n \cdot 2 i \cdot \frac {e^{i \frac \pi n} } {e^{\frac {2 \pi i} n} - 1}\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \frac \pi n \cdot \frac {2 i e^{i \frac \pi n} } {e^{\frac \pi n i} \paren {e^{\frac \pi n i} - e^{-\frac \pi n i} } }\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \frac \pi n \cdot \frac {2 i} {e^{\frac \pi n i} - e^{-\frac \pi n i} }\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \frac \pi n \csc \left({\frac \pi n}\right)\) $\quad$ Cosecant Exponential Formulation $\quad$


$\blacksquare$