# How to compute $H_{5+\sqrt{7}}$

How to calculate $$H_{5+\sqrt{7}}$$ where $$H_n$$ is the nth harmonic number.

If we use the integral representation of harmonic numbers then we have:

$$H_{5+\sqrt{7}}=\int_{0}^{1}\frac{x^{5+\sqrt{7}}-1}{x-1}dx$$

I don't know how to calculate the integral.

But how I can approx the value of $$H_{5+\sqrt{7}}$$?

• I believe $H_x = \psi (x + 1) + \gamma$, where $\psi$ is the digamma function and $\gamma$ is the Euler--Mascheroni constant. There are several ways to compute $\psi$. – Gary Feb 13 at 11:57
• @Gary, is there any way to compute the integral? – user715522 Feb 13 at 11:59
• There are better integral representations, see dlmf.nist.gov/5.9.ii You may use some software to implement them. However, most softwares come with the digamma function implemented in them already. You may also consider Wolfram Alpha. – Gary Feb 13 at 12:06
• What is $n$ in the second expression? – Wojowu Feb 13 at 12:14
• Now your second expression doesn't make sense, since a sum can't go up to $5+\sqrt{7}$. – Wojowu Feb 13 at 12:29

There is numerical value: $$2.675338453513690479902288408923721418916296445960334$$
There is a recurrence formula: $$H_x=H_{x-1}+\frac1x$$, which together with the expansion $$H_x = \frac{\pi^2}6x+\frac{\psi_2(1)}2x^2+\frac{\pi^4}{90}x^3+\ldots$$ and tables for polygamma functions allows you to approximate the harmonic number.
There is a formula for rational approximation of argument: $$H_{\frac {p}{q}}={\frac {q}{p}}+2\sum _{k=1}^{\lfloor {\frac {q-1}{2}}\rfloor }\cos \left({\frac {2\pi pk}{q}}\right)\ln \left({\sin \left({\frac {\pi k}{q}}\right)}\right)-{\frac {\pi }{2}}\cot \left({\frac {\pi p}{q}}\right)-\ln \left(2q\right)}$$
• Could you add that this works for $p<q$ ? – Claude Leibovici Feb 14 at 16:24