Integration #5

Let $f(x) = \displaystyle \prod_{k=1}^n (x+k)$ , then:

$\begin{aligned} f(x) & = (x+1)(x+2)(x+3)\cdots(x+n) \\ f'(x) & = (x+2)(x+3)(x+4)\cdots(x+n) + (x+1)(x+3)(x+4)\cdots(x+n) + (x+1)(x+2)(x+4)\cdots(x+n) + \cdots + (x+1)(x+2)(x+3)\cdots(x+n-1) \\ & = \frac {\prod_{k=1}^n (x+k)}{x+1} + \frac {\prod_{k=1}^n (x+k)}{x+2} + \frac {\prod_{k=1}^n (x+k)}{x+3} + \cdots + \frac {\prod_{k=1}^n (x+k)}{x+n} \\ & = \prod_{k=1}^n (x+k) \sum_{r=1}^n \frac 1{x+r} \end{aligned}$

Therefore,

$\begin{aligned} \int_0^1 \prod_{k=1}^n (x+k) \sum_{r=1}^n \frac 1{x+r} dx & = \int_0^1 f'(x) \ dx \\ & = f(1) - f(0) \\ & = \prod_{k=1}^n (k+1) - \prod_{k=1}^n k \\ & = (n+1)! - n! \\ & = n!(n+1-1) \\ & = \boxed{n\cdot n!} \end{aligned}$

I will construct a quick argument via integration by parts, but it needs some setup. Let $P_n (x)= \prod_{k=1}^n (x+k)$ and $S_n (x) = \sum_{r=1}^n \frac{1}{x+r}$ . Notice that the anti-derivative of $S_n (x)$ is $\sum_{r=1}^n \log(x+r) = \log \prod_{r=1}^n (x+r) = \log P_n (x)$ . Another important observation is that $P_n(0) = n!$ and $P_n(1) = (n+1)!$ . With that, we want to compute:

$\int_0^1 P_n(x) S_n(x) dx$ . Let $u = P_n(x), du = (P_n(x))' dx$ and $dv = S_n(x)dx, v = \log P_n (x)$ so by integration by parts this is equal to: $P_n(x) \log P_n(x) \bigg|_0^1 - \int_0^1 \log(P_n(x)) (P_n(x))' dx = (n+1)! \log((n+1)!) - n! \log(n!) - \int_0^1 \log(P_n(x)) (P_n(x))' dx$ .

I will now just focus on the integral. Let $u = P_n(x)$ . Then the integral is equal to $\int_{n!}^{(n+1)!} \log u du = u \log u - u \bigg|_{n!}^{(n+1)!} = (n+1)! \log((n+1)!) - (n+1)! - n! \log(n!) + n!$ .

Adding that to the whole result we get $(n+1)! - n! = n! ((n+1) - 1) = n! n$ .