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digamma
digamma (daɪˈgæmə) [a. L. digamma, Gr. δίγαµµα the digamma, f. δι- twice + γάµµα the letter gamma: so called by the grammarians of the first century, from its shape ϝ or F, resembling two gammas (Γ) set one above the other.] The sixth letter of the original Greek alphabet, corresponding to the Semit... Oxford English Dictionary
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Digamma function
The digamma function is often denoted as or (the uppercase form of the archaic Greek consonant digamma meaning double-gamma). Special values The digamma function has values in closed form for rational numbers, as a result of Gauss's digamma theorem. wikipedia.org
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Series involving Digamma relates to Exponential Integral I came across the following series involving the Digamma function $\Psi$: \begin{equation} \sum^{\infty}_{k=0} \Psi(k+1) \frac{z^k}{k!}, \end{equation} where z...
Here is a derivation based on $$\psi(k+1)=-\gamma+\sum_{n=1}^{k}\frac{1}{n}=-\gamma+\int_0^1\frac{1-t^k}{1-t}\,dt$$ (let me use the conventional notation) and the formula $$\gamma=\int_0^1\frac{1-e^{-x}}{x}\,dx-\int_1^\infty\frac{e^{-x}}{x}\,dx.$$ We get \begin{align}\sum_{k=0}^{\infty}\psi(k+1)\fra...
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A property of the _digamma_ function On the last paragraph of the article on Wikipedia about the digamma function I found 1 _The digamma function appears in the regularization of divergent integrals_ $$ \int _{0}^{\i...
The idea of a regularization is to subtract some (constant) infinity to be left with a finite expression possibly meaning something. In this case, the relevant formula is mentioned in the article at Wikipedia: $$\psi (a)=-\gamma +\sum _{n=0}^{\infty }\left({\frac {1}{n+1}}-{\frac {1}{n+a}}\right).$$...
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Digamma series Representation I was reading about where the series representation of digamma is proved and its states: $$-\frac{1}{x}-\gamma +\sum_{n=1}^{+\infty}\left(\frac{1}{n}-\frac{1}{n+x}\right) = -\gamma +\sum...
A good starting point is the Weierstrass product for the $\Gamma$ function: $$ \Gamma(x) = \frac{e^{-\gamma x}}{x}\prod_{n\geq 1}\left(1+\frac{x}{n}\right)^{-1}e^{x/n}\tag{1} $$ By considering $\frac{d}{dx}\log(\cdot)$ of both sides, $$ \psi(x) = -\frac{1}{x}-\gamma+\sum_{n\geq 1}\frac{x}{n(n+x)}=-\...
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Integral representation of the Digamma function The digamma function is defined to be $\psi^{(0)}(x)=\frac{d}{dx}ln(\Gamma(x))$, from which we can derive: $$\psi^{(0)}(x)=\frac{d}{dx}ln(\Gamma(x))=\frac{\Gamma'(x)}{\G...
As a rule of thumb, it is best to apply a logarithmic derivative to a product. For instance, if we start with the Weierstrass product for the $\Gamma$ function $$ \Gamma(z+1) = e^{-\gamma z}\prod_{n\geq 1}\left(1+\frac{z}{n}\right)^{-1}e^{z/n}\tag{1} $$ we instantly get: $$ \psi(z+1) = -\gamma+\sum_...
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Newton Series for the Digamma Function The Wikipedia page for the digamma function states that the following holds: $$\psi(s+1) = -\gamma - \sum_{k=1}^\infty \frac{(-1)^k}{k}\binom{s}{k}.$$ Is there any textbook/paper...
See the result in the display immediately before the words "third Retkes identity" at the English Wikipedia page for Harmonic number. It is the derivation $H_n = \dots = \sum_{k=1}^{n}(-1)^{k-1} \frac{1}{k} \binom{n}{k}$, which gives you the less obvious piece in the identity you see.
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gamma函数一一digamma函数的求导 - 知乎 - 知乎专栏
1、什么是gamma函数伽玛函数（Gamma Function）作为 阶乘函数（严格上是f(x)=(x-1)!)的延拓，是定义在复数范围内的亚纯函数，通 常写成 \Gamma(x) ，负整数和0是它的一阶极点。（1）在实数域上gamma函数定义为： \…
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Why is the digamma function written as ψ rather than ϝ or Ϝ? The gamma function is written with the upper-case letter Γ, but the digamma function is usually written with the lower-case letter ψ. How did this notation ...
For the sake of marking this as "answered", this question was addressed on the History of Mathematics forum: this function was used, and denoted “psi”, much before it got a name.
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Sum of complex digamma functions It seems that the sum of the digamma function of $z$ and the digamma function of its conjugate $z^*$ is always real-valued. $$\psi(z)+\psi(z^*)=\frac{\Gamma'(z)}{\Gamma(z)}+\frac{\Gamm...
$\psi(z) - \overline{\psi(\overline{z})}$ is analytic (except perhaps at the singularities of $\psi$) and is $0$ on the positive real axis (because $\psi$ is real there), so by analytic continuation it must be $0$ everywhere. Thus $\psi(z) + \psi(\overline{z}) = \psi(z) + \overline{\psi(z)}$ is alwa...
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How to prove Gauss's Digamma Theorem? Here $\psi(z)$ is digamma function, $\Gamma(z)$ is gamma function. $$\psi(z)=\frac{{\Gamma}'(z)}{\Gamma(z)},$$ For positive integers $m$ and $k$ (with $m < k$), the digamma functi...
You can look at this, and the references therein. **Added** : In fact, a quick Google search gives several references for the proof. Also, if the math does not render well, the Planetmath team suggests to switch the view style to HTML with pictures (you can choose at the bottom of the page).
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Difficult Limit involving digamma function Evaluate: $$\lim_{z \to 0} \psi(-z)\cdot \bigg ( 1 - 2z(z+1) \bigg) - z\cdot\psi'(-z) $$ If we simply substitute in $0$ that gets us infinity, and problems. The answer is $...
Since we can write this limit as a residue, an answer is already given here. Anyway, since in a punctured neighbourhood of the origin we have: $$ \psi(-x)=\frac{1}{x}-\gamma+\zeta(2)x+\ldots, $$ $$-\psi'(-x) = -\frac{1}{x^2}+\zeta(2)+\ldots,$$ it follows that: $$ (1-2x(1+x))\,\psi(-x) = -\frac{1}{x}...
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How to Derive this Digamma identity? !enter image description here I dont see the transition from $(-z)^k$ in the fist sum to the transition to $(z+2)^k$ in the second sum? How is that derived?
From $$\psi(1-z)=-\gamma-\sum_{k\geq 1}\zeta(k+1) z^{k} $$ and $\psi(2-z)=\psi(1-z)+\frac{1}{1-z}$ it follows that: $$\psi(2-z) = \frac{1}{1-z}-\gamma-\sum_{k\geq 1}\zeta(k+1) z^{k}=(1-\gamma)+\sum_{k\geq 1}(1-\zeta(k+1))\,z^k\tag{1}$$ now replace $z$ with $z+2$ to get the stated identity.
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Trigonometric term in digamma function $\psi_{0}(-n)$ Solutions to expressions s.a. $$ S(n)=\sum_{k=1}^{n}\frac{1}{k-r} = \psi_{0}(n-r+1)- \psi_{0}(1-r), $$ involves digamma function. For positive values it has the...
It comes, I think, from the formula $$ \pi \cot(\pi z) = \frac{1}{z} + \sum_{k=1}^\infty \frac{2z}{z^2-k^2} = \frac{1}{z} + \sum_{k=1}^\infty \left( \frac{1}{k+z} - \frac{1}{k-z}\right) $$
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An integral related to the digamma function I was playing around with asymptotics and integral formulas of the digamma function and exponential integral when I stumbled upon this one: $$\psi(x)=\int_0^\infty\left(\fr...
It is not clear what you meant with "not using $\psi(x)$". For $\Re(x) > 0$ $$\int_0^\infty\left(\frac{e^{-t}}t-\frac{e^{-xt}}{1-e^{-t}}\right)dt = \int_0^\infty e^{-t}\left(\frac{1}t-\frac{1}{1-e^{-t}}\right)dt+ \int_0^\infty\left(\frac{e^{-t}-e^{-xt}}{1-e^{-t}}\right)dt$$ $$ = C+\sum_{n=0}^\infty ...
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