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Find $f(n,r)$ in the below equation;$$\int \frac{1}{\prod_{r=0}^{n} (x+r)}\, dx=\sum_{r=0}^n \frac{(-1)^r}{f(n,r)}p(x)+K$$

Here, K is the constant of Integration.

What I have tried so far;

$$L=\int \frac{1}{\prod_{r=0}^{n} (x+r)}\, dx=\int \frac{1}{(x)(x+1)(x+2)(x+3)...(x+n)}\,dx$$

The first thought that came to me was evaluating L using partial fractions;

$$\frac{1}{\prod_{r=0}^{n} (x+r)}=\frac{t_1}{x}+\frac{t_2}{(x+1)}+\frac{t_3}{(x+2)}+\cdots+\frac{t_{n+1}}{(x+n)}=\sum_{k=0}^n \frac{t_{k+1}}{(x+k)}$$

But I am not able to proceed from here, it would be great to see how to solve this, for reference the answer is $f(n,r)=r!(n-r)!$

NadiKeUssPar
  • 2,474

1 Answers1

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You can show that the partial fraction expansion is

$\dfrac1{\prod_{k=0}^n (x-k)} =\sum_{k=0}^{n} \dfrac{(-1)^{n-k}}{(x-k)k!(n-k)!} =\dfrac1{n!}\sum_{k=0}^{n} \dfrac1{x-k}(-1)^{n-k}\binom{n}{k} $.

Write $\dfrac1{\prod_{k=0}^n (x-k)} =\sum_{k=0}^n \dfrac{a_k}{x-k} $, multiply both sides by $\prod_{k=0}^n (x-k) $, and set $x=k$ for $k=0$ to $n$ to get the $a_k$.

marty cohen
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