Editing Harmonic number (section)

===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>