双函数gydF4y2Ba

的gydF4y2Ba双函数gydF4y2Ba通常用gydF4y2Ba $\ψgydF4y2Ba$定义为?的对数导数gydF4y2Baγ函数gydF4y2Ba。gydF4y2Ba

$\ psi (z) = \ dfrac {\ mathrm {d}} {\ mathrm z d {}} \ ln \γ(z) = \ dfrac{\伽马^ \ ' (z)}{\γ(z)}gydF4y2Ba$

$\ psi (s + 1) = \ psi (s) + \ dfrac{1}{年代}gydF4y2Ba$

考虑gydF4y2Ba

$\γ(s + 1) = s \伽马(s)。gydF4y2Ba$

取自然对数gydF4y2Ba

$\ ln \大型γ(s + 1)(\ \大)= \ ln \大型γ(s)(\ \大)+ \ ln (s)。gydF4y2Ba$

关于的微分gydF4y2Ba $年代,gydF4y2Ba$我们得到了gydF4y2Ba

$\ psi (s + 1) = \ psi (s) + \ dfrac{1}{}。\ _ \广场gydF4y2Ba$

$\ psi (s + 1) = - \伽马+ \ sum_ {k = 1} ^ \ infty \离开(\ dfrac {1} {k} - \ dfrac {1} {k + s} \右)gydF4y2Ba$

考虑函数的Weierstrass表示:gydF4y2Ba

$\γ(s) = \ dfrac {e ^{- \伽马年代}}{年代}\ prod_ {k = 1} ^ \ infty e ^ {s / k} \离开(1 + \ dfrac{年代}{k} \右)^{1}。gydF4y2Ba$

取自然对数,gydF4y2Ba

$\ ln \大型γ(s)(\ \大)= - \伽马s - \ ln (s) + \ sum_ {k = 1} ^ \ infty \离开(\ dfrac{年代}{k} - \ ln \大(1 + \ dfrac{年代}{k} \大)\右)。gydF4y2Ba$

关于的微分gydF4y2Ba $年代,gydF4y2Ba$我们有gydF4y2Ba

$\ psi (s) = - \γ- \ dfrac{1}{年代}+ \ sum_ {k = 1} ^ \ infty \离开(\ dfrac {1} {k} - \ dfrac {1} {k + s} \右)= - \伽马+ \ sum_ {k = 1} ^ \ infty \离开(\ dfrac {1} {k} - \ dfrac {1} {k + s - 1} \右)。gydF4y2Ba$

因此,取代gydF4y2Ba $年代gydF4y2Ba$与gydF4y2Ba $s + 1gydF4y2Ba$给了gydF4y2Ba

$\ psi (s + 1) = - \伽马+ \ sum_ {k = 1} ^ \ infty \离开(\ dfrac {1} {k} - \ dfrac {1} {k + s} \右)。\ _ \广场gydF4y2Ba$

$\ psi (s + 1) = - \伽马+ \ int_0 ^ 1 \ dfrac {1 - x ^年代}{1 - x} dxgydF4y2Ba$

考虑gydF4y2Ba

$\开始{对齐}\ psi (s + 1) & = - \伽马+ \ sum_ {n = 1} ^ \ infty \离开(\ dfrac {1} {n} - \ dfrac {1} {n + s} \) \ \ & = -γ+ \ \ sum_ {n = 1} ^ \ infty \ int_0 ^ 1 \大(x ^ {n} - x ^ {n + s - 1} \大)dx \ \ & = -γ+ \ \ int_0 ^ 1 \ sum_ {n = 1} ^ \ infty \大(x ^ {n} - x ^ {n + s - 1} \大)dx。结束\{对齐}gydF4y2Ba$

通过应用gydF4y2Ba几何级数求和gydF4y2Ba,我们有gydF4y2Ba

$\ psi (s + 1) = - \伽马+ \ int_0 ^ 1 \ dfrac {1 - x ^年代}{1 - x} dx。\ _ \广场gydF4y2Ba$

由此，我们可以很容易地找到双伽玛函数的具体值;例如,把gydF4y2Ba $s = 0,gydF4y2Ba$我们得到了gydF4y2Ba $\ psi(1) = - \γ。gydF4y2Ba$

同样，通过积分表示gydF4y2Ba谐波数gydF4y2Ba，gydF4y2Ba

$\psi(s+1) = -\gamma + H_s。gydF4y2Ba$

由欧拉反射公式可得:gydF4y2Ba

$\psi(1-z) - \psi(z) = \pi \cot \pi z。gydF4y2Ba$

欧拉反射公式如下:gydF4y2Ba

$\Gamma(z)\Gamma(1-z) = \frac{\pi}{\sin \pi z}。gydF4y2Ba$

取自然对数，对上述表达式求导，我们观察到gydF4y2Ba

$大(\ \ ln \γ(z) \大)+ \ ln \大型γ(1 - z)(\ \大)=罪\ log \π- \ log \ \πz。gydF4y2Ba$

微分，我们得到gydF4y2Ba

$\开始{对齐}\ dfrac{\伽马^ {\ '}(z)}{\γ(z)} - \ dfrac{\伽马^ {\ '}(1 - z)}{\γ(1 - z)} & = - \ dfrac{\π\因为\πz}{\罪\πz} \ \ \ \ \ psi (z) - \ psi (1 - z) & = - \ dfrac{\π\因为\πz}{\罪\πz} \ \ \ \ \ psi (1 - z) - \ psi (z) & z = \π\床\π\{对齐}结束gydF4y2Ba$

在哪里gydF4y2Ba $\ψgydF4y2Ba$表示gydF4y2Ba双函数gydF4y2Ba的对数导数gydF4y2Baγ函数gydF4y2Ba。gydF4y2Ba $_ \广场gydF4y2Ba$

$2 \ψ(2 s) = 2 \ ln(2) + \ψ(s) + \ \离开(s + \ frac12 \右)gydF4y2Ba$

考虑gydF4y2Ba

$大概{\π}\ \ \γ(2 s) = 2 ^ {2 s - 1} \γ(s) \γ\ (s + \ frac12 \右)。gydF4y2Ba$

日志,gydF4y2Ba

$大(\ \ ln \ sqrt{\π}\大)+ \ ln \大型γ(2 s)(\ \大)= (2 s - 1) \ ln (2) + \ ln \大型γ(s)(\ \大)+ \ ln \离开(\伽马\大(s +{\小\ frac12} \大)\右)。gydF4y2Ba$

关于的微分gydF4y2Ba $年代,gydF4y2Ba$我们有gydF4y2Ba

$2 \ψ(2 s) = 2 \ ln(2) + \ψ(s) + \ \ (s + \ frac12 \右)。\ _ \广场gydF4y2Ba$

证明gydF4y2Ba

$\ sum_ {n = 0} ^ \ infty \ dfrac {1} {n ^ 2 + 1} = \ dfrac {\ pi + 1} {2} + \ dfrac{\π}{e ^{2 \π}1}。gydF4y2Ba$

---

我们有gydF4y2Ba

$S = \ sum_ {n = 0} ^ {\ infty} \压裂{1}{n ^ 2 + 1} = \压裂{1}{2我}\ sum_ {n = 0} ^ {\ infty} \离开(\压裂{1}{n} - \压裂{1}{n +我}\右)。gydF4y2Ba$

重写这个和式得到gydF4y2Ba

$\开始{对齐}2 & = \ sum_ {n = 1} ^ {\ infty} \离开(\压裂{1}{n-1-i} - \压裂{1}{n - 1 + i} \) \ \ & = \ sum_ {n = 1} ^ {\ infty} \离开(\压裂{1}{n} - \压裂{1}{n - 1 + i} \右)- \ sum_ {n = 1} ^ {\ infty} \离开(\压裂{1}{n} - \压裂{1}{n-1-i} \右)。结束\{对齐}gydF4y2Ba$

通过双伽函数的级数表示，这就是gydF4y2Ba

$\开始{对齐}2 & = \ psi (i) - \ psi(我)\ \ & = \ psi (i) - \ psi(我)- \ dfrac{1}{我}\ \ & = - \π\床(我\π)+我\ \ & = \π\双曲余切(\π)+我。结束\{对齐}gydF4y2Ba$

进一步简化，我们得到gydF4y2Ba

$S = \ dfrac{1 + \π\双曲余切(\π)}{2}\意味着S = \ dfrac {\ pi + 1} {2} + \ dfrac{\π}{e ^{2 \π}1}。\ _ \广场gydF4y2Ba$

的gydF4y2Ba $n ^ \文本{th}gydF4y2Ba$多函数由gydF4y2Ba

$\ psi_n (s) = \ dfrac {d ^ n} {ds ^ n} \ psi (s) = \ psi ^ {(n)}(年代)。gydF4y2Ba$

我们可以从中得到很多性质;例如，通过微分级数表示gydF4y2Ba $ngydF4y2Ba$次,我们有gydF4y2Ba

$\ psi_n (s) = (1) ^ {n + 1} n !\ sum_ {k = 1} ^ \ infty \ dfrac {1} {(k + s - 1) ^ {n + 1}} = (1) ^ {n + 1} n !\ sum_ {k = 0} ^ \ infty \ dfrac {1} {(k + s) ^ {n + 1}} = (1) ^ {n + 1} n ! \泽塔(n + 1, s),gydF4y2Ba$

在哪里gydF4y2Ba $\泽塔(n + 1, s)gydF4y2Ba$是gydF4y2Ba赫维茨ζ函数gydF4y2Ba。把gydF4y2Ba $s = 1,gydF4y2Ba$我们可以得到gydF4y2Ba $\ psi_n (1) = (1) ^ {n + 1} n ! \泽塔(n + 1)。gydF4y2Ba$

由此，我们还可以得到狄格函数的泰勒级数:gydF4y2Ba

${对齐}\ \开始psi (s) & = \ sum_ {n = 0} ^ \ infty \ dfrac {\ psi ^ {(n)} (1) (s - 1) ^ n} {n !} \ \ & = -γ+ \ \ sum_ {n = 1} ^ \ infty \ dfrac {\ psi_n (1) (s - 1) ^ n} {n !} \ \ & = - \γ- \ sum_ {n = 1} ^ \ infty (1) ^ n \泽塔(n + 1) (s - 1) ^ n \ \ \ psi (s + 1) & = - \γ- \ sum_ {n = 1} ^ \ infty \泽塔(n + 1) (- s) ^ n。结束\{对齐}gydF4y2Ba$

我们可以对积分表示求导gydF4y2Ba $ngydF4y2Ba$次获得gydF4y2Ba

$\ psi_n (s + 1) = \ int_0 ^ 1 \ dfrac {\ ln ^ n (x) x ^年代}{x - 1} dx。gydF4y2Ba$

我们也可以对泛函方程这样做得到gydF4y2Ba

$\ psi_n (s + 1) = \ psi_n (s) + (1) ^ nn !z ^ {n - 1}。gydF4y2Ba$

digamma函数可以在Mathematica中实现如下:gydF4y2Ba

1gydF4y2Ba

多[z]gydF4y2Ba

或gydF4y2Ba

1gydF4y2Ba

多[0,z]gydF4y2Ba