6
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ i.e. you want to directly show that$$\int_0^1{1-x^\alpha\over1-x}{\rm~d}x=\sum_{k=1}^\infty\frac\alpha{k(k+\alpha)}$$? $\endgroup$ – Simply Beautiful Art Dec 26 '17 at 19:25
  • $\begingroup$ @SimplyBeautifulArt yes that's correct! $\endgroup$ – alfalfa Dec 26 '17 at 19:25
9
$\begingroup$

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 $$H_{\alpha} = \int_{0}^{1} \lim_{n\to\infty} f_n (x) \, dx = \lim_{n\to\infty} \int_{0}^{1} f_n(x) \, dx = \sum_{k=1}^{\infty} \frac{\alpha}{k(k+\alpha)}$$

$\endgroup$
  • $\begingroup$ thank you very much, the extra rigor is very helpful $\endgroup$ – alfalfa Dec 27 '17 at 15:55
7
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ ahh i had basically done this but just didn't see how close i was. Thanks a lot $\endgroup$ – alfalfa Dec 26 '17 at 19:33
  • $\begingroup$ No problem $\ddot\smile$ $\endgroup$ – Simply Beautiful Art Dec 26 '17 at 19:34
3
$\begingroup$

We start with

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

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.

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

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.