# Prove series form of fractional harmonic numbers

Let $H_\alpha$ be the $\alpha$th fractional harmonic number so that $$H_\alpha = \int_0^1 \frac{1-x^\alpha}{1-x}\,\text dx.$$

I want to directly show $$H_\alpha = \sum_{k=1}^\infty \frac{\alpha}{k(k+\alpha)}.$$

I know this must be true because $\psi(1 + \alpha) = -\gamma + H_\alpha$ and $\psi(1 + \alpha) = -\gamma + \sum_{k=1}^\infty \frac{\alpha}{n(n+\alpha)}$ where $\psi$ is the digamma function but I haven't managed to prove it.

I've proven this for $\alpha \in \mathbb N$ because in this case $(1-x^\alpha)/(1-x) = \sum_{i=0}^{\alpha-1}x^i$ and I can split up $\frac{\alpha}{k(k+\alpha)} = \frac{1}{k} - \frac{1}{k+\alpha}$ to get telescoping since eventually $1/k' = 1/(k+\alpha)$. But if $\alpha \notin \mathbb N$ then I get neither the factorization nor the telescoping so it seems those tricks only help in $\mathbb N$. Even if $\alpha = p/q \in \mathbb Q$ the telescoping fails and I don't see any way to generalize my approach in $\mathbb N$, so it seems an entirely different approach may be needed. So how can I prove this directly? I've also tried a few different series for $\frac{1-x^\alpha}{1-x}$ but no luck so far.

I'm sure there are many ways to show this using fancy properties of $\psi$ and other special functions, but I'm trying to prove it directly. I'm only bringing up $\psi$ for context. Thanks a lot for any help.

• i.e. you want to directly show that$$\int_0^1{1-x^\alpha\over1-x}{\rm~d}x=\sum_{k=1}^\infty\frac\alpha{k(k+\alpha)}$$? – Simply Beautiful Art Dec 26 '17 at 19:25
• @SimplyBeautifulArt yes that's correct! – alfalfa Dec 26 '17 at 19:25

We have $$H_{\alpha} = \int_{0}^{1} \frac{1 - x^{\alpha}}{1-x} \, dx = \int_{0}^{1} \frac{1}{1-x} - \frac{x^{\alpha}}{1-x} \, dx = \int_{0}^{1} \lim_{n\to\infty} \sum_{k=0}^{n} (x^k - x^{k+\alpha}) \, dx$$ Setting $f_n (x) = \sum_{k=0}^{n} x^k - x^{k+\alpha}$, we see $$\int_{0}^{1} f_n (x) \, dx = \sum_{k=0}^{n} \left(\frac{1}{k+1} - \frac{1}{k+1+\alpha}\right) = \sum_{k=0}^{n} \frac{\alpha}{(k+1)(k+1+\alpha)} = \sum_{k=1}^{n} \frac{\alpha}{k(k+\alpha)}$$ Note that $f_{n+1} (x) = x^{n+1}(1-x^{\alpha}) + f_n (x)$, and thus for $x\in [0,1]$ we have $f_{n+1} (x) \ge f_n (x)$. Hence, by the monotone convergence theorem, we may conclude $$H_{\alpha} = \int_{0}^{1} \lim_{n\to\infty} f_n (x) \, dx = \lim_{n\to\infty} \int_{0}^{1} f_n(x) \, dx = \sum_{k=1}^{\infty} \frac{\alpha}{k(k+\alpha)}$$

• thank you very much, the extra rigor is very helpful – alfalfa Dec 27 '17 at 15:55

Hint:

$$\frac{1-x^\alpha}{1-x}=(1-x^\alpha)\sum_{k=0}^\infty x^k=\sum_{k=0}^\infty(x^k-x^{k+\alpha})$$

Integrate termwise (with appropriate justification) and adjust the indices (so it start with $k=1$) and you should be done.

• ahh i had basically done this but just didn't see how close i was. Thanks a lot – alfalfa Dec 26 '17 at 19:33
• No problem $\ddot\smile$ – Simply Beautiful Art Dec 26 '17 at 19:34

$$H_\alpha = \int_{0}^1 \frac{1-x^\alpha}{1-x}\tag{1}$$

valid for $Re(\alpha \gt -1$.

Partial integration and expanding the $\log$ gives

$$\begin{array} &H_\alpha&= -\log(1-x) (1-x^\alpha)|_{x=0}^{x=1} -\alpha \int_{0}^1 x^{\alpha-1} \log(1-x) \, dx\\ &=-\alpha \int_{0}^1 x^{\alpha-1} \log(1-x)\, dx\\ &=\alpha \int_{0}^1 x^{\alpha-1} \sum_{k\ge 1}\frac{x^k}{k}\, dx\\ &=\alpha \sum_{k\ge 1} \frac{1}{k} \int_{0}^1 x^{k+\alpha-1}\, dx\\ &=\alpha \sum_{k\ge 1} \frac{1}{k(k+\alpha)}\tag{2}\\ \end{array}$$

Notice that (1) and this derivation (because we want the partially integrated part to vanish) is valid only for $Re(\alpha) \gt -1$ but (2) which can also be written as

$$H_\alpha = \sum_{k\ge 1}(\frac{1}{k}-\frac{1}{k+\alpha})\tag{3}$$

gives the analytic continuation of $H_\alpha$ to arbitrary complex $\alpha$.

(3) shows that the only singularities of $H_\alpha$ are simple poles at the negative integers.

qed.

• thanks for the answer, i appreciate the different approach – alfalfa Dec 27 '17 at 15:54