It is well known that the $n$-th harmonic number $H_n$ has the integral representation $\int_0^1 \frac{1-x^n}{1-x}$. If we replace $n$ with rational non-integer $p$, do we ever get a rational outcome?
My only thought, due to W|A, is that $$\int\frac{1-x^p}{1-x}=-\frac{x^{p+1}}{p+1} {}_2F_1(1,p+1;p+2; x) - \log(1-x),$$ where $ {}_2F_1$ is the hypergeometric function, so what we're looking at is$$\lim_{x\to1^-}-\frac{1}{p+1} {}_2F_1(1,p+1; p+2; x)-\log(1-x).$$
As noted by whacka in the comments, the integral is equal to $$p\sum_{n=1}^\infty \frac{1}{n(n+p)}=\gamma + \psi^{(0)}(1+p),$$ (where $\gamma$ is the Euler-Mascheroni constant) which was proved transcendental by M. Ram Murty and N. Saradha in 2007.
Note that in the link there is a typo, actually $1\le a\le q-1$, and that using $$ \psi^{(0)}\left(1+ \frac aq \right) = \psi^{(0)}\left(\frac aq\right) + \frac qa$$ we have transcendence for larger values of $a$ too (excluding multiples of $q$, of course).