Hurwitz-zeta function

Integral representation of the Hurwitz-zeta function

We show the proof of the following result posted by @integralsbot: $$\int_{0}^{1} \frac{x^n \ln^{a-1} \frac{1}{x} }{1-x} dx = \Gamma(a) \left( \zeta (a)- H_{n}^{(a)} \right)$$ This result is the integral representation of the Hurwitz-zeta function. Recall that: $$\zeta(a,n+1) = \left( \zeta (a)- H_{n}^{(a)} \right)$$ The proof for the case $a \in \mathbb{R}$ is well known, so here we show a proof for $a \in \mathbb{N}$ using the Feynman's trick just for the sake to give a different proof.

Proof

For $\displaystyle a\in \mathbb{N}$
$$\begin{align*} \int_{0}^{1} \frac{x^n \ln^{a-1} \frac{1}{x} }{1-x} dx = & \int_{0}^{1} \frac{x^n \left(\frac{d^{a-1}}{dt^{a-1}}\Big|_{t=0+} \frac{1}{x^t}\right) }{1-x} dx\\ =& \frac{d^{a-1}}{dt^{a-1}}\Big|_{t=0+}\int_{0}^{1} \frac{x^{n-t} }{1-x} dx\\ =& \frac{d^{a-1}}{dt^{a-1}}\Big|_{t=0+}\int_{0}^{1} \sum_{k=0}^{\infty}x^kx^{n-t} dx\\ =& \frac{d^{a-1}}{dt^{a-1}}\Big|_{t=0+} \sum_{k=0}^{\infty} \int_{0}^{1}x^{k+n-t} dx\\ =& \frac{d^{a-1}}{dt^{a-1}}\Big|_{t=0+} \sum_{k=0}^{\infty} \frac{1}{(k+n+1-t)} \\ =& \sum_{k=0}^{\infty} \left[\frac{d^{a-1}}{dt^{a-1}}\Big|_{t=0+} \frac{1}{(k+n+1-t)} \right] \\ =& \sum_{k=0}^{\infty} \left[\lim_{t \to 0+} (a-1)! \frac{1}{(k+n+1-t)^a} \right] \\ =& (a-1)! \sum_{k=0}^{\infty} \frac{1}{(k+n+1)^a} \\ =& (a-1)! \zeta(a,n+1) \\ =& \Gamma(a) \left( \zeta (a)- H_{n}^{(a)} \right) \\ \end{align*}$$ $$\boxed{ \int_{0}^{1} \frac{x^n \ln^{a-1} \frac{1}{x} }{1-x} dx =\Gamma(a) \left( \zeta (a)- H_{n}^{(a)} \right) }$$