Prove series form of fractional harmonic numbers

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

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.

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.