<s>[INST] Closed-form of $\int_0^1 B_n(x)\psi(x+1)\,dx$ Is there a closed-form of the following integral?
$$I_n = \int_0^1 B_n(x)\psi(x+1)\,dx,$$
where $B_n(x)$ are the Bernoulli polynomials and $\psi(x)$ is the digamma function.
The motivation of the problem was this question. In this answer achille hui showed that for all $n>0$
$$\int_0^1 B_n\left(\left\{ \frac1x \right\}\right) \frac{dx}{x}
= -\int_0^1 B_n(u) \psi(1+u) du,$$
where $\{x\}$ is the fractional part of $x$.
By looking at specific $n$ values we could observe interesting patterns.
$$\begin{align}
I_0 & =  0\\
I_1 & =  1-\frac 12 \ln(2\pi) \\
I_2 & =  -\frac 12 + 2 \ln A \\
I_3 & = \frac{1}{12} - \frac{3}{4\pi^2}\zeta(3) = \frac{1}{12} - \frac{1}{8}\frac{\zeta(3)}{\zeta(2)}\\
I_4 & = \frac{1}{45} - 4\zeta'(-3)\\
I_5 & = -\frac{13}{360} + \frac{15}{4\pi^4}\zeta(5) = -\frac{13}{360} + \frac{1}{24}\frac{\zeta(5)}{\zeta(4)}\\
I_6 & = -\frac{1}{252}-6\zeta'(-5)\\
I_7 & = \frac{47}{1260} - \frac{315}{8\pi^6}\zeta(7) = \frac{47}{1260} - \frac{1}{24}\frac{\zeta(7)}{\zeta(6)}\\
I_8 & = -\frac{8}{1575}-8\zeta'(-7)\\
I_9 & = -\frac{1703}{25200} -\frac{2835}{4\pi^8}\zeta(9) = -\frac{1703}{25200} -\frac{3}{40}\frac{\zeta(9)}{\zeta(8)}\\
I_{10} & = \frac{2461}{83160} - 10\zeta'(-9)\\
\dots
\end{align}$$
where $A$ is the Glaisher–Kinkelin constant, $\zeta(3)$ is the Apéry's constant and in general $\zeta$ is the Riemann zeta function and $\zeta'$ is its derivative.
 [/INST] This is an answer for the odd case.
Proposition. Let $k=1,2,3,\ldots$. Then

$$
\int_0^1 B_{2k+1}(x)\: \psi (x+1) \:dx=(-1)^{k+1}\frac{(2k+1)!}{(2\pi)^{2k+1}}\pi \: \zeta(2k+1)-\sum_{j=0}^{2k}\!\frac{ {{2k+1}\choose j} B_j}{2k+1-j} \quad (*)
$$

The proof is here, observing that
$$\begin{align}
\int_0^1 B_{2k+1}(x)\: \psi (x+1) \:dx & =  \int_0^{1} \frac{B_{2k+1}\left(\left\{ 1/x \right\}\right)}{x}dx. \tag1
\end{align}$$ A proof of $(1)$ may be found here.
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