Editing Harmonic number (section)

==Harmonic numbers for real and complex values==
{{unreferenced section|date=May 2019}}
The formulae given above,
<math display="block"> H_x = \int_0^1 \frac{1-t^x}{1-t} \, dt= \sum_{k=1}^\infty {x \choose k} \frac{(-1)^{k-1}}{k}</math>
are an integral and a series representation for a function that interpolates the harmonic numbers and, via [[analytic continuation]], extends the definition to the complex plane other than the negative integers ''x''. The interpolating function is in fact closely related to the [[digamma function]]
<math display="block">H_x = \psi(x+1)+\gamma,</math>
where {{math|''ψ''(''x'')}} is the digamma function, and {{math|''γ''}} is the [[Euler–Mascheroni constant]]. The integration process may be repeated to obtain
<math display="block">H_{x,2}= \sum_{k=1}^\infty \frac {(-1)^{k-1}}{k} {x \choose k} H_k.</math>

The [[Taylor series]] for the harmonic numbers is
<math display="block">H_x=\sum_{k=2}^\infty (-1)^{k}\zeta (k)\;x^{k-1}\quad\text{ for } |x| < 1</math>
which comes from the Taylor series for the digamma function (<math>\zeta </math> is the [[Riemann zeta function]]).

=== Alternative, asymptotic formulation ===
There is an asymptotic formulation that gives the same result as the analytic continuation of the integral just described. When seeking to approximate&nbsp;{{math|''H''{{sub|''x''}}}} for a [[complex number]]&nbsp;{{math|''x''}}, it is effective to first compute&nbsp;{{math|''H''{{sub|''m''}}}} for some large integer&nbsp;{{math|''m''}}.  Use that as an approximation for the value of&nbsp;{{math|''H''{{sub|''m''+''x''}}}}. Then use the recursion relation {{math|1=''H''{{sub|''n''}} = ''H''{{sub|''n''−1}} + 1/''n''}} backwards&nbsp;{{math|''m''}} times, to unwind it to an approximation for&nbsp;{{math|''H''{{sub|''x''}}}}.  Furthermore, this approximation is exact in the limit as&nbsp;{{math|''m''}} goes to infinity.

Specifically, for a fixed integer&nbsp;{{math|''n''}}, it is the case that
<math display="block">\lim_{m \rightarrow \infty} \left[H_{m+n} - H_m\right] = 0.</math>

If&nbsp;{{math|''n''}} is not an integer then it is not possible to say whether this equation is true because we have not yet (in this section) defined harmonic numbers for non-integers.  However, we do get a unique extension of the harmonic numbers to the non-integers by insisting that this equation continue to hold when the arbitrary integer&nbsp;{{math|''n''}} is replaced by an arbitrary complex number&nbsp;{{math|''x''}},

<math display="block">\lim_{m \rightarrow \infty} \left[H_{m+x} - H_m\right] = 0\,.</math>
Swapping the order of the two sides of this equation and then subtracting them from&nbsp;{{math|''H''<sub>''x''</sub>}} gives
<math display="block">
\begin{align}H_x &= \lim_{m \rightarrow \infty} \left[H_m - (H_{m+x}-H_x)\right] \\[6pt]
&= \lim_{m \rightarrow \infty} \left[\left(\sum_{k=1}^m \frac{1}{k}\right) - \left(\sum_{k=1}^m \frac{1}{x+k}\right) \right] \\[6pt]
&= \lim_{m \rightarrow \infty} \sum_{k=1}^m \left(\frac{1}{k} - \frac{1}{x+k}\right) = x \sum_{k=1}^{\infty} \frac{1}{k(x+k)}\, .
\end{align}
</math>

This [[infinite series]] converges for all complex numbers&nbsp;{{math|''x''}} except the negative integers, which fail because trying to use the recursion relation {{math|1=''H''{{sub|''n''}} = ''H''{{sub|''n''−1}} + 1/''n''}} backwards through the value&nbsp;{{math|1=''n'' = 0}} involves a division by zero.  By this construction, the function that defines the harmonic number for complex values is the unique function that simultaneously satisfies (1) {{math|1=''H''{{sub|0}} = 0}}, (2) {{math|1=''H''{{sub|''x''}} = ''H''{{sub|''x''−1}} + 1/''x''}} for all complex numbers&nbsp;{{math|''x''}} except the non-positive integers, and (3) {{math|1=lim<sub>''m''→+∞</sub> (''H''<sub>''m''+''x''</sub> − ''H''<sub>''m''</sub>) = 0}} for all complex values&nbsp;{{math|''x''}}.

This last formula can be used to show that
<math display="block"> \int_0^1 H_x \, dx = \gamma, </math>
where&nbsp;{{math|''γ''}} is the [[Euler–Mascheroni constant]] or, more generally, for every&nbsp;{{math|''n''}} we have:
<math display="block"> \int_0^nH_{x}\,dx = n\gamma + \ln(n!) .</math>

===Special values for fractional arguments===
There are the following special analytic values for fractional arguments between 0 and 1, given by the integral
<math display="block">H_\alpha = \int_0^1\frac{1-x^\alpha}{1-x}\,dx\, .</math>

More values may be generated from the recurrence relation
<math display="block"> H_\alpha = H_{\alpha-1}+\frac{1}{\alpha}\,,</math>
or from the reflection relation
<math display="block"> H_{-\alpha}-H_{\alpha-1} = \pi\cot{(\pi\alpha)}.</math>

For example:
<math display="block"> \begin{align}
H_{\frac{1}{2}} &= 2 - 2\ln 2 \\
H_{\frac{1}{3}} &= 3 - \frac{\pi}{2\sqrt{3}} - \frac{3}{2}\ln 3 \\
H_{\frac{2}{3}} &= \frac{3}{2}+\frac{\pi}{2\sqrt{3}} - \frac{3}{2}\ln 3 \\
H_{\frac{1}{4}} &= 4 - \frac{\pi}{2} - 3\ln 2 \\
H_{\frac{1}{5}} &= 5 - \frac{\pi}{2} \sqrt{1+\frac{2}{\sqrt{5}}} - \frac{5}{4} \ln 5 - \frac{\sqrt{5}}{4} \ln\left(\frac{3+\sqrt{5}}{2}\right) \\
H_{\frac{3}{4}} &= \frac{4}{3} + \frac{\pi}{2} - 3\ln 2 \\
H_{\frac{1}{6}} &= 6 - \frac{\sqrt{3}}{2} \pi - 2\ln 2 - \frac{3}{2} \ln 3 \\
H_{\frac{1}{8}} &= 8 - \frac{1+\sqrt{2}}{2} \pi - 4\ln{2} - \frac{1}{\sqrt{2}} \left(\ln\left(2 + \sqrt{2}\right) - \ln\left(2 - \sqrt{2}\right)\right) \\
H_{\frac{1}{12}} &= 12 - \left(1+\frac{\sqrt{3}}{2}\right)\pi - 3\ln{2} - \frac{3}{2} \ln{3} + \sqrt{3} \ln\left(2-\sqrt{3}\right)
\end{align}</math>

Which are computed via [[Digamma function#Gauss's digamma theorem|Gauss's digamma theorem]], which essentially states that for positive integers ''p'' and ''q'' with ''p'' < ''q''
<math display="block"> H_{\frac{p}{q}} = \frac{q}{p} +2\sum_{k=1}^{\lfloor\frac{q-1}{2}\rfloor} \cos\left(\frac{2 \pi pk}{q}\right)\ln\left({\sin \left(\frac{\pi k}{q}\right)}\right)-\frac{\pi}{2}\cot\left(\frac{\pi p}{q}\right)-\ln\left(2q\right)</math>

===Relation to the Riemann zeta function===
Some derivatives of fractional harmonic numbers are given by
<math display="block">
\begin{align}
\frac{d^n H_x}{dx^n} & = (-1)^{n+1}n!\left[\zeta(n+1)-H_{x,n+1}\right] \\[6pt]
\frac{d^n H_{x,2}}{dx^n} & = (-1)^{n+1}(n+1)!\left[\zeta(n+2)-H_{x,n+2}\right] \\[6pt]
\frac{d^n H_{x,3}}{dx^n} & = (-1)^{n+1}\frac{1}{2}(n+2)!\left[\zeta(n+3)-H_{x,n+3}\right].
\end{align}
</math>

And using [[Taylor series|Maclaurin series]], we have for ''x'' < 1 that
<math display="block">
\begin{align}
H_x & = \sum_{n=1}^\infty (-1)^{n+1}x^n\zeta(n+1) \\[5pt]
H_{x,2} & = \sum_{n=1}^\infty (-1)^{n+1}(n+1)x^n\zeta(n+2) \\[5pt]
H_{x,3} & = \frac{1}{2}\sum_{n=1}^\infty (-1)^{n+1}(n+1)(n+2)x^n\zeta(n+3).
\end{align}
</math>

For fractional arguments between 0 and 1 and for ''a'' > 1,
<math display="block">
\begin{align}
H_{1/a} & = \frac{1}{a}\left(\zeta(2)-\frac{1}{a}\zeta(3)+\frac{1}{a^2}\zeta(4)-\frac{1}{a^3} \zeta(5) + \cdots\right) \\[6pt]
H_{1/a, \, 2} & = \frac{1}{a}\left(2\zeta(3)-\frac{3}{a}\zeta(4)+\frac{4}{a^2}\zeta(5)-\frac{5}{a^3} \zeta(6) + \cdots\right) \\[6pt]
H_{1/a, \, 3} & = \frac{1}{2a}\left(2\cdot3\zeta(4)-\frac{3\cdot4}{a}\zeta(5)+\frac{4\cdot5}{a^2}\zeta(6)-\frac{5\cdot6}{a^3}\zeta(7)+\cdots\right).
\end{align}
</math>