TITLE: On Harmonic numbers at half-integer values
QUESTION [5 upvotes]: 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?

REPLY [5 votes]: 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
$$2\int_{0}^{1}\frac{x^{2n+2}-1}{x^2-1}\,dx -2\log(2)=2\sum_{k=0}^{n}\frac{1}{2k+1}-2\log(2)=2H_{2n+1}-H_n-2\log(2). $$