Why does $\Psi(n)=H_{n-1}-\gamma$

Question

According to wikipedia: The gamma function obeys the equation

$\Gamma(z+1)=z\Gamma(z)$.

Taking the derivative with respect to $z$ gives:

$\Gamma'(z+1)=z\Gamma'(z)+\Gamma(z)$

Dividing by $\Gamma(z+1)$ or the equivalent $z\Gamma(z)$ gives:

$\frac{\Gamma'(z+1)}{\Gamma(z+1)}=\frac{\Gamma'(z)}{\Gamma(z)}+\frac{1}{z}$

or:

$\psi(z+1)=\psi(z)+\frac{1}{z}$

Since the harmonic numbers are defined as

$H_n=\sum_{k=1}^n\frac{1}{k}\tag{1}$

the digamma function is related to it by:

$\psi(n)=H_{n-1}-\gamma$

where $H_n$ is the nth harmonic number, and $\gamma$ is the Euler–Mascheroni constant.

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This doesn't make sense to me. The two values subtracted appear to be equal: How does $$\psi(z+1)=\psi(z)+\frac{1}{z}\tag{2}$$ differ from $\sum_1^n \frac{1}{z}=H_n$ If the sum in (2) doesn't terminate at 1, the definition of the sum is undefined for gamma at negative integer values. If wikipedia is saying non integer values of $\Psi$ of a finite sum i.e. $\frac{1}{1.5}+\frac{1}{2.5}+\frac{1}{3.5}\dots$ differ from $H_n$ by $\gamma$ I'm equally doubtful as $\gamma$ is only equal to infinite sums and I don't see how this value could equal the infinite difference between the integral and sum $\frac{1}{n}$. What am I not understanding. Please help.

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Edit: as noted in the comments, could someone explain/link/prove why $\frac{\Gamma'(0)}{\Gamma(0)}=\gamma$

Answer 1

You have $$\psi(n) = \psi(n-1)+\frac{1}{n-1} = \psi(n-2)+\frac{1}{n-2}+\frac{1}{n-1} = \dotsb = \psi(1) + \sum_{k=1}^{n-1} \frac{1}{k}. $$ So what is $\psi(1)?$ Obviously it's $\Gamma'(1)$. We have the useful identity $$ I_n(z) := \int_0^n t^{z-1}\left( 1-\frac{t}{n} \right)^n dt = \frac{n^z n!}{z(z+1)\dotsm(z+n)}. $$ Since the left-hand side tends to $\Gamma(z)$ for $z>0$, we have Gauss's definition $$ \Gamma(z) = \lim_{n \to \infty} \frac{n^z n!}{z(z+1)\dotsm(z+n)} $$ (In fact, this definition holds for any complex $z$ for which the denominator makes sense, but never mind about that.) We have $$ \frac{I_n'(z)}{I_n(z)} + \frac{1}{z+n} = \log{n} - \sum_{k=1}^{n} \frac{1}{z+k-1}. $$ Putting $z = 1$ and taking $n \to \infty$ gives $\Gamma'(1)/1 = -\gamma$.

(Worry about interchanging the limits can be removed by doing the calculation more carefully, but it's basically just using the Monotone or Dominated Convergence Theorem on the integral.)

Answer 2

Differentiating the log of the identity $\Gamma(x+1)=x\Gamma(x)$, we get $$ \frac{\Gamma'(x+1)}{\Gamma(x+1)}=\frac1x+\frac{\Gamma'(x)}{\Gamma(x)}\tag{1} $$ which, because $H_x=\frac1x+H_{x-1}$, implies that, for some constant $C$, $$ \frac{\Gamma'(x+1)}{\Gamma(x+1)}=C+H_x\tag{2} $$ Since $\log(\Gamma(x+1))-\log(\Gamma(x))=\log(x)$, we have $$ \lim_{x\to\infty}\left(\frac{\Gamma'(x+1)}{\Gamma(x+1)}-\log(x)\right)=0\tag{3} $$ Equation $(2)$, limit $(3)$, and $\lim\limits_{x\to\infty}(H_x-\log(x))=\gamma$ imply that $C=-\gamma$. Therefore, $$ \frac{\Gamma'(x+1)}{\Gamma(x+1)}=-\gamma+H_x\tag{4} $$

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It is not true that $\frac{\Gamma'(0)}{\Gamma(0)}=\gamma$. I think what you're thinking about is $$ \frac{\Gamma'(1)}{\Gamma(1)}=-\gamma+H_0=-\gamma\tag{5} $$

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Additional Reference

In this answer it is shown that $$ \frac{\Gamma'(x)}{\Gamma(x)}=-\gamma+\sum_{k=0}^\infty\left(\frac1{k+1}-\frac1{k+x}\right)\tag{6} $$ and in this answer, it is shown that $$ H_x=\sum_{k=1}^\infty\left(\frac1k-\frac1{k+x}\right)\tag{7} $$ Combining $(6)$ and $(7)$ gives $(4)$.

Answer 3

The Euler-Mascheroni constant $\gamma$ comes in most easily from the equivalent definition for the Gamma function $$\frac{1}{\Gamma(s)}= se^{\gamma s}\prod_{n=1}^{\infty} \left(1+\frac{s}{n}\right)e^{-s/n}$$ In turn that expression can be shown to be equivalent to Euler's definition as follows:

$$\begin{align} \lim_{n\to\infty}\frac{(n-1)!n^s}{s(s+1)\cdots(s+n-1)}& = \frac{1}{s}\lim_{n\to\infty}\left[n^s\prod_{k=1}^{n-1}\left(\frac{k}{s+k}\right)\right] \\ &=\frac{1}{s}\lim_{n\to\infty}\left[e^{s\log n}\prod_{k=1}^{n-1}\left(1+\frac sk\right)^{-1}\right]\\ & =\frac{1}{s}\lim_{n\to\infty}\left[\exp\left(s\log n-s\sum_{m=1}^{n-1}\frac1m\right)\prod_{k=1}^{n-1}\left(1+\frac sk\right)^{-1}e^{s/k}\right] \\ & =\frac{1}{s}e^{-s\gamma}\prod_{k=1}^{\infty}\left(1+\frac sk\right)^{-1}e^{s/k} \end{align} $$ This and a number of other basic results about the Gamma function can be found in some old typewritten notes by Bruce Berndt (Rudiments of the theory of the gamma function) which can be found floating around the internet.

Finally to obtain the formula for the logarithmic derivative take the $\log$ (principal branch) $$\log \Gamma(s) = \sum_{n=1}^{\infty}\left[\frac sn-\log\left(1+\frac sn\right)\right] - \log s - \gamma s$$ and take the derivative $$\frac{\Gamma'(s)}{\Gamma(s)}=\sum_{n=1}^{\infty}\left(\frac{1}{n}-\frac{1}{n+s}\right) - \frac{1}{s}-\gamma$$ Setting $s=1$ and evaluating the telescoping sum gives the desired formula.