Closed form for fixed $m$ to $\int\frac{dx}{x(x+1)(x+2)(x+3)...(x+m)}$

Question

$I=\displaystyle\int\frac{dx}{x(x+1)(x+2)(x+3)...(x+m)}$

Attempt:

$\dfrac{ A_0 }{ x }+\dfrac{ A_1 }{ x +1 }+\dfrac{ A_2 }{ x + 2 }...+\dfrac{ A_m }{ x +m } =\dfrac{1}{x(x+1)(x+2)(x+3)...(x+m)}$

But things got very messy.

I also thought that 1)applying integral by parts, or 2) taking terms one from left head, one from right hand and use some kind of a symmetry, or 3) using trigonometric identites etc.

I cannot see the solution, any hint, help would be perfect. Thank you in advanced.

Answer 1

Starting from your attempt, notice that for $k \in \{0, \cdots, m\}$ we have

$$ A_k = \lim_{x\to-k} \frac{x+k}{x(x+1)\cdots(x+m)} = \frac{(-1)^k}{k!(m-k)!} = \frac{(-1)^k}{m!}\binom{m}{k}. $$

So we have

$$ \frac{1}{x(x+1)\cdots(x+m)} = \frac{1}{m!} \sum_{k=0}^{m} \binom{m}{k} \frac{(-1)^k}{x+k} $$

and now you can integrate term by term.

Answer 2

Maybe the following reduction will be helpful: $$x(x+1)(x+2)...(x+(m-2))(x+(m-1))(x+m)=x(x+m)(x+1)(x+(m-1))(x+2)(x+m(m-1))...=(x^2+mx)(x^2+mx+1(m-1))(x^2+mx+2(m-2))$$

Answer 3

$f(x) = \frac {1}{x(x+1)(x+2)\cdots(x+m)} = \frac {A_0}{x} + \cdots + \frac {A_n}{x+m}\\ \lim_\limits{x\to -n} (x-n)f(x) = A_n\\ A_n = \prod_\limits {i\ne n} \frac 1{i-n}$