Integral Representation of Dirichlet Beta Function in terms of Gamma Function
Jump to navigation
Jump to search
Theorem
- $\ds \map \beta s = \frac 1 {\map \Gamma s} \int_0^\infty \frac {x^{s - 1} e^{-x} } {1 + e^{-2 x} } \rd x$
where:
- $\beta$ denotes the Dirichlet beta function
- $\Gamma$ denotes the gamma function
- $s$ is a complex number with $\map \Re s > 0$.
Proof
We have, by Laplace Transform of Power:
- $\ds \frac {\paren {-1}^n \map \Gamma s} {\paren {2 n + 1}^s} = \paren {-1}^n \int_0^\infty x^{s - 1} e^{-\paren {2 n + 1} x} \rd x$
for $\map \Re s > 0$.
Summing, we have:
| \(\ds \map \Gamma s \sum_{n \mathop = 0}^N \frac {\paren {-1}^n} {\paren {2 n + 1}^s}\) | \(=\) | \(\ds \sum_{n \mathop = 0}^N \paren {-1}^n \int_0^\infty x^{s - 1} e^{-\paren {2 n + 1} x} \rd x\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \int_0^\infty x^{s - 1} \sum_{n \mathop = 0}^N \paren {-1}^n e^{-\paren {2 n + 1} x} \rd x\) | Linear Combination of Definite Integrals | |||||||||||
| \(\ds \) | \(=\) | \(\ds \int_0^\infty x^{s - 1} e^{-x} \paren {\sum_{n \mathop = 0}^N \paren {-e^{-2 x} }^n} \rd x\) |
We have:
| \(\ds \lim_{N \mathop \to \infty} \paren {\map \Gamma s \sum_{n \mathop = 0}^N \frac {\paren {-1}^n} {\paren {2 n + 1}^s} }\) | \(=\) | \(\ds \map \Gamma s \lim_{N \mathop \to \infty} \paren {\sum_{n \mathop = 0}^N \frac {\paren {-1}^n} {\paren {2 n + 1}^s} }\) | Multiple Rule for Limits of Complex Functions | |||||||||||
| \(\ds \) | \(=\) | \(\ds \map \Gamma s \map \beta s\) | Definition of Dirichlet Beta Function |
Therefore:
| \(\ds \map \Gamma s \map \beta s\) | \(=\) | \(\ds \lim_{N \mathop \to \infty} \paren {\int_0^\infty x^{s - 1} e^{-x} \paren {\sum_{n \mathop = 0}^N \paren {-e^{-2 x} }^n} \rd x}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \int_0^\infty x^{s - 1} e^{-x} \paren {\lim_{N \mathop \to \infty} \sum_{n \mathop = 0}^N \paren {-e^{-2 x} }^n} \rd x\) | Lebesgue's Dominated Convergence Theorem | |||||||||||
| \(\ds \) | \(=\) | \(\ds \int_0^\infty x^{s - 1} e^{-x} \paren {\sum_{n \mathop = 0}^\infty \paren {-e^{-2 x} }^n} \rd x\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \int_0^\infty \frac {x^{s - 1} e^{-x} } {1 - \paren {-e^{- 2x} } } \rd x\) | Sum of Infinite Geometric Sequence | |||||||||||
| \(\ds \) | \(=\) | \(\ds \int_0^\infty \frac {x^{s - 1} e^{-x} } {1 + e^{-2 x} } \rd x\) |
giving:
- $\ds \map \beta s = \frac 1 {\map \Gamma s} \int_0^\infty \frac {x^{s - 1} e^{-x} } {1 + e^{-2 x} } \rd x$
$\blacksquare$