0
$\begingroup$

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 $H_{5+\sqrt{7}}$?

$\endgroup$
  • $\begingroup$ 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 $\psi$. $\endgroup$ – Gary Feb 13 at 11:57
  • $\begingroup$ @Gary, is there any way to compute the integral? $\endgroup$ – user715522 Feb 13 at 11:59
  • $\begingroup$ 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. $\endgroup$ – Gary Feb 13 at 12:06
  • 1
    $\begingroup$ What is $n$ in the second expression? $\endgroup$ – Wojowu Feb 13 at 12:14
  • $\begingroup$ Now your second expression doesn't make sense, since a sum can't go up to $5+\sqrt{7}$. $\endgroup$ – Wojowu Feb 13 at 12:29
0
$\begingroup$

What answer do you want?

There is numerical value: $2.675338453513690479902288408923721418916296445960334$

There is a method how to approximate the integrals.

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: $$ {\displaystyle H_{\frac {p}{q}}={\frac {q}{p}}+2\sum _{k=1}^{\lfloor {\frac {q-1}{2}}\rfloor }\cos \left({\frac {2\pi pk}{q}}\right)\ln \left({\sin \left({\frac {\pi k}{q}}\right)}\right)-{\frac {\pi }{2}}\cot \left({\frac {\pi p}{q}}\right)-\ln \left(2q\right)} $$

Choose any.

$\endgroup$
  • $\begingroup$ Could you add that this works for $p<q$ ? $\endgroup$ – Claude Leibovici Feb 14 at 16:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy