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Harmonic numbers are usually defined, for $n\in\Bbb N$, by $$H_n=\sum_{k=1}^{n}\frac1k$$ But then one may note, $$H_n=\sum_{k=1}^{n}\int_0^1x^{k-1}\mathrm dx=\int_0^1\frac{1-x^n}{1-x}\mathrm dx=\int_0^1\frac{x^n-1}{x-1}\mathrm dx$$ Which provides somewhat of an extension to real $n>-1$. Hence I define $$H(n)=\int_0^1\frac{x^n-1}{x-1}\mathrm dx$$ And I begin on finding a general expression for $H\big(\frac{2n+1}2\big)$ for $n\in\Bbb N\cup\{-1,0\}$.

I start by defining the polynomial $P_k(x)$ for $k\in \Bbb N$ by the property $$x^n-1=(x-1)P_k(x)$$ It is fairly easily shown that $$P_k(x)=\sum_{m=0}^{k-1}x^m$$ Which leads us to the great fact that $$H_n=\int_0^1P_n(x)\mathrm dx$$ With this in mind we begin the integral $$H\bigg(\frac{2n+1}2\bigg)=\int_0^1\frac{x^{(2n+1)/2}-1}{x-1}\mathrm dx$$ we make the substitution $x=u^2$: $$H\bigg(\frac{2n+1}2\bigg)=2\int_0^1\frac{u^{2n+1}-1}{u^2-1}u\mathrm du$$ $$H\bigg(\frac{2n+1}2\bigg)=2\int_0^1P_{2n+1}(u)\frac{u\mathrm du}{u+1}$$ $$H\bigg(\frac{2n+1}2\bigg)=2\int_0^1P_{2n+1}(u)\bigg(1-\frac{1}{u+1}\bigg)\mathrm du$$ $$H\bigg(\frac{2n+1}2\bigg)=2\int_0^1P_{2n+1}(u)\mathrm du-2\int_0^1P_{2n+1}(u)\frac{\mathrm du}{u+1}$$ $$H\bigg(\frac{2n+1}2\bigg)=2H_{2n+1}-2\int_0^1P_{2n+1}(u)\frac{\mathrm du}{u+1}$$

Using The series for $P_k$, we have $$ \begin{align} \int_0^1P_{2n+1}(x)\frac{\mathrm dx}{x+1}=&\int_0^1\sum_{i=0}^{2n}\frac{x^i}{x+1}\mathrm dx\\ =&\sum_{i=0}^{2n}\int_0^1\frac{x^i}{x+1}\mathrm dx\\ =&\int_0^1\frac{\mathrm dx}{x+1}+\sum_{i=1}^{2n}\int_0^1\frac{x^i}{x+1}\mathrm dx\\ =&\log2+\sum_{i=1}^{2n}\int_0^1\frac{x^i}{x+1}\mathrm dx\\ =&\log2+? \end{align} $$

I do not know what the best approach for this final integral would be. Could I have some help?

clathratus
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    Why don't you use the standard extension $H_x=\psi(x)+\gamma?$ – gammatester Dec 11 '18 at 18:41
  • @gammatester I didn't know that was a thing. – clathratus Dec 11 '18 at 18:44
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    Small correction after edit timeout: $H_x=\psi(x+1)+\gamma$ – gammatester Dec 11 '18 at 18:48
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    $\displaystyle H:\mathbb{C}\setminus\left{-1,-2,-3,\ldots\right} \to \mathbb{C}$ such that $\displaystyle H_{z} = \Psi\left(z + 1\right) + \gamma$. $\displaystyle \Psi$ is the $\displaystyle Digamma\ Function$ and $\displaystyle\gamma$ is the $\displaystyle Euler$-$\displaystyle Mascheroni\ Constant$. – Felix Marin Dec 12 '18 at 05:29

1 Answers1

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In general, by setting $\psi(x)=\frac{d}{dx}\log\Gamma(x)$, the Weierstrass product for the $\Gamma$ function ensures $$ H_n = \psi(n+1) + \gamma $$ for any $n\in\mathbb{N}$, with $\gamma$ being the Euler-Mascheroni constant. Legendre's duplication formula for $\Gamma$ gives a duplication formula for the $\psi$ (digamma) function too, from which it follows that

$$ H_{n+\frac{1}{2}} = \psi\left(n+\frac{3}{2}\right)+\gamma=\int_{0}^{1}\frac{x^{n+1/2}-1}{x-1}\,dx=2\int_{0}^{1}\frac{x^{2n+2}-x}{x^2-1}\,dx $$ equals $$2\int_{0}^{1}\frac{x^{2n+2}-1}{x^2-1}\,dx -2\log(2)=2\sum_{k=0}^{n}\frac{1}{2k+1}-2\log(2)=2H_{2n+1}-H_n-2\log(2). $$

Jack D'Aurizio
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  • Great! Thank you. – clathratus Dec 11 '18 at 18:58
  • 1+ Excellent work. I tried to treat $H_{n+\frac{1}{3}}$ along the same lines and found $H_{n+\frac{1}{3}}=H_{3 (n+1)}+2 \Re\left(e^{\frac{i \pi }{3}} B_{e^{\frac{2 i \pi }{3}}}(3 n+4,0)\right)-\log (3)$ where $B$ is the incomplete Beta function but I'm not sure that this is the equivalent to the result for the case $n+\frac{1}{2}$. On the other hand we know that $H(3x)=\frac{1}{3}(H_x+H_{x−1/3}+H_{x−2/3})−\log(3)$ but I can't find $H_{n+1/3}$ from it. Could you give me a hint? Mille grazie – Dr. Wolfgang Hintze Feb 15 '23 at 08:21
  • For complex $n\to x$ we have $H_{x+\frac{1}{3}}=H_{3 (x+1)}-2 \Re\left(e^{2 i \pi \left(x+\frac{1}{3}\right)} B_{e^{-\frac{1}{3} (2 i \pi )}}(3 x+4,0)\right)-\log (3)$ – Dr. Wolfgang Hintze Feb 15 '23 at 10:25