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

Question

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

Answer 1

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.

Answer 2

I don't think this is mathematically correct, but it'll still provide a good approximation. A (fairly) smooth curve of $\operatorname{H}(x)$ seems to be defined by: $$\operatorname{H}\left(x\right)=\frac{\operatorname{mod}\left(x,1\right)}{x}+\operatorname{H}\left(\lfloor{x}\rfloor\right)$$ for all real $x\geq 1$, where $\lfloor x\rfloor$ is the floor function. For the purpose of $\operatorname{mod}\left(x,1\right)$ gives the fractional part of $x$, and the division by $x$ allows the growth rate of that portion of the equation to decay just as quickly as that of $\operatorname{H}\left(\lfloor{x}\rfloor\right)$. Even if this isn't correct, it'll give the decent approximation of $2.67731598605..$ which is exactly $\frac{363}{140}+\frac{\operatorname{mod}\left(\sqrt{7},1\right)}{\sqrt{7}}$.