Definite Integral to Infinity of Reciprocal of 1 plus Power of x
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:
This article, or a section of it, needs explaining. In particular: $\theta \nearrow \dfrac \pi 2$ needs to be explained, as it does not exist anywhere else on $\mathsf{Pr} \infty \mathsf{fWiki}$. You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by explaining it. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Explain}} from the code. |
\(\ds 2 \int_0^{\pi / 2} \paren {\tan \theta}^{\frac 2 n - 1} \rd \theta\) | \(=\) | \(\ds 2 \int_0^\infty \paren {\tan \theta}^{\frac 2 n - 1} \frac {\rd x} {\frac 2 n \paren {1 + \paren {\tan^{\frac 2 n} \theta}^n} \cdot \paren {\tan \theta}^{\frac 2 n - 1} }\) | 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:
- $\ds C_R = \set {R e^{i \theta} : 0 \le \theta \le \frac {2 \pi} n}$
This article, or a section of it, needs explaining. In particular: the fact that we are working in the complex plane. This is not immediately obvious. You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by explaining it. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Explain}} from the code. |
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.
This article, or a section of it, needs explaining. In particular: Diagram You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by explaining it. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Explain}} from the code. |
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 |
$\blacksquare$