Definite Integral to Infinity of Reciprocal of 1 plus Power of x/Proof 1
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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
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:
- $\ds \int_0^\infty \frac 1 {1 + x^n} \rd x = \frac \pi n \map \csc {\frac \pi n}$
$\blacksquare$