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Denote by $B_n(x)=\sum_{k=0}^n\binom{n}{k}B^-_{n-k}x^k$ the $n$-th Bernoulli polynomial, where $B^-_0=1,B^-_1=-\frac{1}{2},B^-_2=\frac{1}{6},...$ are the Bernoulli numbers. Im interested in describing the polynomial $\overset{\sim}{B}_n$ given by $$ \overset{\sim}{B}_n(x):=\sum_{k=0}^n\binom{n}{k}B^-_{n-k}H_kx^k $$ where $H_k = \sum_{j=1}^k\frac{1}{j}$ are the harmonic numbers. Is there a nice recursion for the sequence of polynomials $(\overset{\sim}{B}_n)_{n\geq 0}$? Or even a somewhat explicit formula?

A part from that, I'm also particularly interested in determining $\overset{\sim}{B}_n(1)$.

Here is my approach:

As $H_k=\int_0^1\frac{1-u^k}{1-u}du$, we have $$ \overset{\sim}{B}_n(x)=\sum_{k=0}^n\binom{n}{k}B^-_{n-k}x^k\int_0^1\frac{1-u^k}{1-u}du = \int_0^1\sum_{k=0}^n\binom{n}{k}B^-_{n-k}x^k\frac{1-u^k}{1-u}du=\\ =\int_0^1\frac{B_n(x)-B_n(xu)}{1-u}du=\int_0^x\frac{B_n(x)-B_n(u)}{x-u}du $$ so in particular, we find $$ \overset{\sim}{B}_n(1)=\int_0^1\frac{B_n(1)-B_n(u)}{1-u}du=\int_0^1\frac{B_n(1)-B_n(1-u)}{u}du=\\ =(-1)^n\int_0^1\frac{B_n(0)-B_n(u)}{u}du=(-1)^n\left(\left[(B_n(0)-B_n(u))\log(u)\right]_0^1+\int_0^1B_{n}'(u)\log(u)du\right)=\\ =(-1)^nn\int_0^1B_{n-1}(u)\log(u)du. $$ So is there maybe a way to calculate $\int_0^1B_{n-1}(u)\log(u)du$? I tried to do some more partial integrations, but things got pretty messy.

Redundant Aunt
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    It appears that asymptotically $\int_0^1 B_m(u)\log(u) du \sim (-1)^m \ m!/(2\pi)^m \ C_m$ where $C_{even} = 0.4514116667901...$ and $C_{odd} = 0.7759291740995...$ The form was suggested by using the Fourier series representation for the Bernoulli polynomials, integrating, using an asymptotic form for the sine integral, summing the series in terms of the zeta function, which is asymptotically 1. I don't have a closed or integral form for $C_m,$ just numerics. – user321120 Aug 08 '19 at 16:53
  • A bit more simple form: $$ \overset{\sim}{B}n(1)=(-1)^{n}\frac{ n}{n-1}+\sum _{k=1}^n \frac{(-1)^{k+1} B{n-k} \binom{n}{k}}{k},\quad n\ge2. $$ – Andrew Aug 16 '19 at 19:59
  • The polynomials satisfy the equation $$ \overset{\sim}{B'}n(x)-n\overset{\sim}{B}{n-1}(x)=\frac{B_n(x)-B_n}{x}. $$ It follows that $\overset{\sim}{B'}n(1)=n\overset{\sim}{B}{n-1}(1)$ for $n>1$. – Andrew Aug 17 '19 at 10:20

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