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I was reading through this answer to a question about the lower bound of a least common multiple.

The question is about showing that:

$$\text{lcm}(1,2,\dots,n) > 2^n$$

I was not clear on one step in the answer provided.

Here are the steps that I understand:

(1) I am clear on the definition of $I_{m,n}$ where $1 \le m \le n$:

$$I_{m,n}=\int_0^1x^{m-1}(1-x)^{n-m}dx = \sum_{r=0}^{n-m}\frac{a_r}{m+r}$$ where each $a_r \in \mathbb{Z}$

(2) I am clear that if $\displaystyle\ell_n = \text{lcm}(1,2,\dots,n)$ that:

$$\displaystyle\ell_n I_{m,n} \in \mathbb{Z}$$

I am unclear when Integration by Parts is applied.

(3) The answer states:

Now it can easily be seen that $I_{m,n} = \dfrac{1}{m{n\choose m}}$

Which follow from integration by parts or reduction formulae.

Could someone explain to me how this last step follows from the first two?

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Larry Freeman
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    I'm pretty sure the integral can be written as the beta function, which can be written in terms of the gamma function, which can be written in terms of factorials – Seth Nov 04 '18 at 00:23
  • Thanks for the tip. I'm not familiar with the beta function. I'll review. I am familar with the gamma function. If you can write the equality as the answer, I will be glad to review and accept if it is clear to me. – Larry Freeman Nov 04 '18 at 00:31
  • @seth. Cool stuff: https://en.wikipedia.org/wiki/Beta_function – Larry Freeman Nov 04 '18 at 00:32
  • Thanks. I'll update the question so it is correct and accept your answer. – Larry Freeman Nov 04 '18 at 02:52

1 Answers1

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$I_{m,n}=\int_0^1 x^{m-1}(1-x)^{n-m+1-1}dx=B(m,n-m+1)=\cfrac{\Gamma(m) \Gamma(n-m+1)}{\Gamma(n+1)}$, which, if n and m are positive integers, is equal to $\cfrac{(m-1)!(n-m)!}{n!}=\cfrac{1}{m{n \choose m}}$, as claimed

Seth
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