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Fungrim entry: 661054

logΓ ⁣(1+z)=γz+k=2ζ ⁣(k)k(z)k\log \Gamma\!\left(1 + z\right) = -\gamma z + \sum_{k=2}^{\infty} \frac{\zeta\!\left(k\right)}{k} {\left(-z\right)}^{k}
Assumptions:zC  and  z<1z \in \mathbb{C} \;\mathbin{\operatorname{and}}\; \left|z\right| < 1
\log \Gamma\!\left(1 + z\right) = -\gamma z + \sum_{k=2}^{\infty} \frac{\zeta\!\left(k\right)}{k} {\left(-z\right)}^{k}

z \in \mathbb{C} \;\mathbin{\operatorname{and}}\; \left|z\right| < 1
Fungrim symbol Notation Short description
LogGammalogΓ(z)\log \Gamma(z) Logarithmic gamma function
ConstGammaγ\gamma The constant gamma (0.577...)
Sumnf(n)\sum_{n} f(n) Sum
RiemannZetaζ ⁣(s)\zeta\!\left(s\right) Riemann zeta function
Powab{a}^{b} Power
Infinity\infty Positive infinity
CCC\mathbb{C} Complex numbers
Absz\left|z\right| Absolute value
Source code for this entry:
    Formula(Equal(LogGamma(Add(1, z)), Add(Neg(Mul(ConstGamma, z)), Sum(Mul(Div(RiemannZeta(k), k), Pow(Neg(z), k)), For(k, 2, Infinity))))),
    Assumptions(And(Element(z, CC), Less(Abs(z), 1))))

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2021-03-15 19:12:00.328586 UTC