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Is there a general formula for $$ I_n = \int_0^1 (x-1)(x-2)\dots (x-n) dx $$ in terms of familiar sequences of number (Harmonic numbers, Bell numbers, Bernoulli numbers, etc) ?

Define $$ P_n(x) = \int_0^x (u-1)(u-2)\dots (u-n) du $$ Have the polynomials $P_n(x)$ appeared in the literature? Is there a simple recursive formula for $P_n(x)$? What are some interesting properties of $P_n(x)$?

The explicit formula for $P_n(x)$ can be obtained by expanding $(u-1)(u-2)\dots(u-n)$ and integrating terms by terms $$ P_n(x) = \frac{x^{n+1}}{n+1} + \sum_{k=1}^n \frac{x^{n-k+1}}{n-k+1} \sum_{1 \leq j_1 < \dots < j_k \leq n} (-1)^k j_1 j_2 \dots j_k $$ Also, it is easy to see that $$ P_{n+1}'(x) = (x-n-1)P'_n(x) $$

Pluviophile
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1 Answers1

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In terms of the Stirling numbers of the first kind, $$ (x - 1) (x-2)\cdots (x - n) = \sum\limits_{k = 0}^n {s(n,k)(x - 1)^k } . $$ Thus, $$ I_n = \sum\limits_{k = 0}^n {(-1)^k\frac{{s(n,k)}}{{k + 1}}} . $$ Asymptotically, $$ I_n \sim ( - 1)^{n} \frac{{n!}}{{\log n}} $$ as $n\to +\infty$

Gary
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  • You need to include a factor of $(-1)^k$ in your sum for $I_n$. – Somos May 23 '21 at 13:38
  • Your asymptotic $I_n \sim ( - 1)^{n - 1} \frac{{(n - 1)!}}{{\log ^2 n}}$ is very wrong. It has the wrong sign. Much better is $(-1)^n n!/\log(n)$. – Somos May 26 '21 at 21:15
  • @Somos Thank you for pointing out. That asymptotics was for the signless sum. – Gary May 27 '21 at 05:49