I came across this multivariable integral which states that for any $s\in\mathbb{C}$ and $n>0$, we have the equity below \begin{align} \sum_{k=0}^n (-1)^k \binom{n}{k}(u+k)^{1-s} = (s-1)_n\int_0^1 \cdots \int_0^1 (u+x_1+\cdots+x_n)^{1-s-n} \,dx_1\cdots dx_n \end{align} where $(a)_n=a(a+1)\cdots(a+n-1)$ is the Pochhammer symbol. I have notice that \begin{align} \int_0^1 \cdots \int_0^1 \bigg(u+\sum_{j=1}^n x_j\bigg)^{1-s-n} \,dx_1\cdots dx_n &= \int_0^1 \cdots \int_0^1 \left.\frac{(u+\sum_{j=1}^n x_j)^{2-s-n}}{2-s-n}\right|_{x_1=0}^1 \,dx_2\cdots dx_n\\ &= \frac{1}{s+n-2} \int_0^1 \cdots \int_0^1 \bigg(u+\sum_{j=2}^n x_j\bigg)^{2-s-n}-\bigg(u+1+\sum_{j=2}^n x_j\bigg)^{2-s-n} \,dx_2\cdots dx_n\\ &= \frac{1}{(s+n-3)(s+n-2)}\int_0^1 \cdots \int_0^1 \bigg(u+\sum_{j=3}^n x_j\bigg)^{3-s-n}-2\bigg(u+1+\sum_{j=3}^n x_j\bigg)^{3-s-n}+\bigg(u+2+\sum_{j=3}^n x_j\bigg)^{3-s-n}\,dx_3\cdots dx_n \\ &\vdots \\ &= \frac{1}{(s-1)_n}\bigg(u^{1-s}-n(u+1)^{1-s}+\cdots+(-1)^{n-1}n(u+n-1)^{1-s}+(-1)^n(u+n)^{1-s}\bigg) \end{align} by repeating this process $n$ times. Other than that, I have also try to use multinomial theorem but could not get to anywhere \begin{align} \bigg(u+\sum_{j=1}^n x_j\bigg)^{1-s-n} &= \sum_{m=0}^{1-s-n} \binom{1-s-n}{m}u^{1-s-n-m}\bigg(\sum_{j=1}^n x_j\bigg)^m\\ &= \sum_{m=0}^{1-s-n} \binom{1-s-n}{m}u^{1-s-n-m} \sum_{k_1+\cdots+k_n=m} \frac{m!}{k_1!\cdots k_n!} \prod_{t=1}^n x_t^{k_t} \end{align} I would be appreciated if someone could help me with this with a non-heuristic approach.
1 Answer
We can use operator methods to show the identity is valid. We consider a function $f=f(u)$ and recall the shift operator $E$ and the identity operator $I$ which are defined as \begin{align*} If(u)&=f(u)\\ Ef(u)&=f(u+1) \end{align*} Applying for instance $k$ times the shift operator $E$ to $f(u)$ we obtain \begin{align*} E^kf(u)=E^{k-1}\left(Ef(u)\right)=E^{k-1}f(u+1)=\cdots=f(u+k) \end{align*}
With this notation we can write the binomial sum as \begin{align*} \color{blue}{\sum_{k=0}^n}&\color{blue}{(-1)^k\binom{n}{k}(u+k)^{-s+1}}\\ &=\sum_{k=0}^n(-1)^k\binom{n}{k}E^k\left(u^{-s+1}\right)\\ &\,\,\color{blue}{=\left(I-E\right)^n\left(u^{-s+1}\right)}\tag{1} \end{align*}
Integrating iteratively $n$ times the right hand side we will see, we'll also get (1).
We obtain \begin{align*} &\color{blue}{\left(s-1\right)^{\overline{n}}} \color{blue}{\int_{0}^1\ldots\int_{0}^1\left(u+x_1+\cdots+x_n\right)^{-s-(n-1)}\,dx_1\ldots dx_n}\tag{2.1}\\ &\quad =(-1)\left(s-1\right)^{\overline{n-1}} \int_{0}^1\ldots\int_{0}^1\left.\left(u+x_1+\cdots+x_n\right)^{-s-(n-2)}\right|_{x_1=0}^1\,dx_2\ldots dx_n\tag{2.2}\\ &\quad =(-1)\left(s-1\right)^{\overline{n-1}} \int_{0}^1\ldots\int_{0}^1(E-I)\left(u+x_2+\cdots+x_n\right)^{-s-(n-2)}\,dx_2\ldots dx_n\\ &\quad =(-1)^2\left(s-1\right)^{\overline{n-2}} \int_{0}^1\ldots\int_{0}^1(E-I)^2\left(u+x_3+\cdots+x_n\right)^{-s-(n-3)}\,dx_3\ldots dx_n\\ &\quad=\ldots\\ &\quad =(-1)^{n-1}\left(s-1\right)^{\overline{1}} \int_{0}^1(E-I)^{n-1}\left(u+x_n\right)^{-s}\,dx_n\\ &\quad=(-1)^n(E-I)^{n-1}\left.\left(u+x_n\right)^{-s+1}\right|_{x_n=0}^1\\ &\quad=(-1)^n(E-I)^n\left(u\right)^{-s+1}\\ &\quad\,\,\color{blue}{=(I-E)^n\left(u^{-s+1}\right)} \end{align*} which coincides with (1) showing the claim is valid. We can show the second part using induction to make it more rigorous.
Comment:
In (2.1) we use Don Knuth's rising factorial notation. \begin{align*} s^{\overline{n}}=s(s+1)\cdots(s+n-1) \end{align*}
In (2.2) we have \begin{align*} (s-1)^{\overline{n}}\,\frac{1}{-s-n+2}&=(s-1)s\cdots(s+n-2)\,\frac{(-1)}{s+n-2}\\ &=(-1)\left(s-1\right)^{\overline{n-1}} \end{align*}
