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I got this general formula for $ n\in N$ (I showed it here)

$$\int_0^1 \left(\frac{x}{1-x} \ln x \right)^n dx=n \sum_{p=0}^{n-1}a(n,p+1) (-1)^{n-p} \zeta(p+2)+n! $$ where $a(n,k)$ is the coefficient of $x^{k-1}$ in the expression $$\prod_{p=2}^n (1+xp)$$

I found in A145324 this formula using Stirling numbers of first kind $$ a(n,k)=\sum_{p=0}^{k-1} (-1)^p |s_1(n+1,n-p+1)|$$

MY QUESTION

how can I prove the last formula that I found in OEIS and is there a closed form for it without using series ?

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    $\begingroup$ Just write $\prod_{p=2}^n (1+xp)=x^{n-1}(1/x+n)_{n-1}$ and then use the definition of Stirling numbers of first kind. $\endgroup$
    – Nemo
    Commented Mar 9 at 7:12

1 Answer 1

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From representation $$\prod_{p=2}^n (1+xp) = (-x)^{n+1} (-1/x)_{n+1} (1+x)^{-1}= \sum_{i\geq 0} s_1(n+1,i) (-x)^{n+1-i}\cdot \sum_{j\geq0} (-x)^j, $$ it follows that the coefficient of $x^{k-1}$ in this product equals $$(-1)^{k-1} \sum_{j=0}^{k-1} s_1(n+1,n+2-k+j).$$

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  • $\begingroup$ @CaveJohnson made a related comment earlier. $\endgroup$
    – LSpice
    Commented Mar 9 at 14:52
  • $\begingroup$ is that result is same ? this make it simplest $\endgroup$
    – Faoler
    Commented Mar 9 at 14:54
  • $\begingroup$ @LSpice: It's not quite what I said in my answer. $\endgroup$
    – Max Alekseyev
    Commented Mar 9 at 15:21
  • $\begingroup$ @Faoler: What result do you refer to? $\endgroup$
    – Max Alekseyev
    Commented Mar 9 at 15:21
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    $\begingroup$ @Faoler: It is the same as in your question under the change of variables: $p\mapsto k-1-j$. $\endgroup$
    – Max Alekseyev
    Commented Mar 9 at 19:01

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