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I was told to phrase my question in a more exciting way when I asked it last time. The following is a preliminary consideration. If you don't need it, just scroll down to START HERE. Here we go then:

Imagine you have a really cool integral, \begin{align} I = \int_a^b f(x) \ dx , \end{align} which you wanted to solve numerically. And what do you take for it: A quadrature formula. And one of the most popular quadrature formulas are the Newton-Cotes formulas. Don't be put off, this just means that the supports for my quadrature formula are at a fixed distance from each other. It applies to an interval $[a,b]$: \begin{align*} x_k = a + k \cdot h, \hspace{15pt} h = \frac{b-a}{n}, \hspace{15pt} k = 0,1,\ldots,n \end{align*}

The quadrature error $E_n[f] := I_n - I$ of a Newton-Cotes formula $n$-th degree to $n$ support points, \begin{align*} I = \int_a^b f(x) \, d x, \hspace{30pt} I_n = I_n\left(f,[a,b]\right) = \sum_{k=0}^n \, w_kf(x_k)\,, \end{align*} for a $(n+1)$ times continuously differentiable function $f(x)$ has the representation \begin{align*} E_n[f] &= \frac{f^{(n+1)}(\xi)}{(n+1)!} \int_a^b \omega(x) \,d x, \hspace{30pt} \omega(x) = \prod_{k=0}^n\,(x-x_k) \\ &\leq \frac{1}{(n+1)!} \,\max_{\xi\in[a,b]}\left|f^{(n+1)}(\xi)\right|\,\int_a^b \omega(x) \,d x \end{align*} BUT ALL OF THIS DOESN'T GIVE A SHIT NOW. Because I want to specify the error exactly. It follows: \begin{align*} \int_a^b \omega(x) \,d x &= \int_a^b (x-x_0)(x-x_1)\cdots(x-x_n)\,d x \\ &= \int_a^b (x-x_0)(x-x_0 - h)(x- x_0 - 2h)\cdots(x-x_0-nh)\,d x \\ &= \int_a^b h^{n+1}\,\left(\frac{x-x_0}{h}\right)\left(\frac{x-x_0}{h} - 1\right)\cdots\left(\frac{x-x_0}{h}-n\right)\,d x, \hspace{30pt} \color{red}{u = \frac{x-x_0}{h}} \\ &= h^{n+2} \int_{\frac{a-x_0}{h}}^{\frac{b-x_0}{h}} u(u-1)\cdots(u-n)\,d u \\ &= h^{n+2}\,(n+1)!\,\psi_{n+2}(u)\,\Big|_{u_1=\frac{a-x_0}{h}}^{u_2=\frac{b-x_0}{h}} \\ &= h^{n+2}\,(n+1)!\,\left[\psi_{n+2}(u_2)-\psi_{n+2}(u_1)\right] \end{align*} And please do not run away in panic now. These $\psi_n(u)$ here are Bernoulli polynomials of the second kind, certain weird polynomials like Legendre polynomials, which have great properties that I want to use, see Wikipedia. \begin{align} (n+1)! \cdot \psi_{n+2}(u) = \int u(u-1)\cdots(u-n)\ d u \end{align} For the limits $u_1 = 0$, $u_2 = n$ we find the following expression: \begin{align*} \int_a^b \omega(x) \, d x &= h^{n+2}\,(n+1)!\,\left[\psi_{n+2}(n) - \psi_{n+2}(0)\right] \\ &= h^{n+2}\,(n+1)!\,\left[-|G_{n+2}| - G_{n+2}\right] \color{white}{\frac{1}{2}} \\ &= h^{n+2}\,(n+1)!\,\left[(-1)^n-1\right]\,|G_{n+2}| \color{white}{\frac{1}{2}} \end{align*} where $G_n$ are the Gregory coefficients defined as follows: \begin{align*} \frac{x}{\ln(1+x)} = 1 + \frac{1}{2}x - \frac{1}{12}x^2 + \frac{1}{24}x^3 - \ldots = 1 + \sum_{n=1}^\infty\,G_n\,x^n \end{align*} This means that for even $n$ we cannot find an expression. Thus, for even $n$, we must use a different approximation. For this we double the term $x_{n/2} = \frac{a+b}{2}$. So we calculate in such a way that we consider an additional grid point and thus a double zero for the interpolation polynomial. \begin{align*} \int_a^b \omega(x)\,d x &= h^{n+3} \int_{0}^{n} u(u-1)\cdots(u-\tfrac{n}{2})^2\cdots(u-n)\,d u \\ &= h^{n+3} \bigg[\underbrace{\int_0^{n} u\cdot u(u-1)\cdots(u-n)\,d u}_{\color{red}{=\,I}} - \frac{n}{2}\underbrace{\int_0^{n} u(u-1)\cdots(u-n)\,d u}_{\color{red}{=\,0}}\bigg] \end{align*}

START HERE:

And now we have finally reached the point where I need your help. Your superhuman knowledge of mathematics is now required. Unfortunately, there was probably still no mathematician who has set up a suitable relation for me, which I can use to solve this integral: \begin{align} I = \int_0^{n} u\cdot u(u-1)\cdots(u-n)\,d u \end{align} I want to use this: \begin{align} (n+1)! \cdot \psi_{n+2}(u) = \int u(u-1)\cdots(u-n)\ d u \end{align} I had the following ideas:

  1. Integration by parts:

\begin{align} I &= u\cdot (n+1)!\,\psi_{n+2}(u)\,\Big|_0^{n} - (n+1)! \int_0^{n} \psi_{n+2}(u)\,d u \\ &= (n+1)!\left[n\cdot\psi_{n+2}(n) - \int_0^n \psi_{n+2}(u)\,d u\right] \\ &= -(n+1)!\left[n\cdot|G_{n+2}| + \int_0^n \psi_{n+2}(u)\,d u\right] \end{align} But what is now \begin{align} \int_0^n \psi_{n+2}(u)\,d u \ ? \end{align}

  1. Consider the case $n+1$

\begin{align} (n+2)!\cdot\psi_{n+3}(u) &= \int u(x-1)(u-2)\cdots(u-n)(u-(n+1))\ du \\ &=\underbrace{\int u \cdot u(u-1)(u-2)\cdots(u-n) \ du}_{\color{red}{=\,I(u)}} - (n+1) \underbrace{\int u(u-1)(u-2)\cdots(u-n) \ du}_{\color{red}{=\,(n+1)!\,\cdot\,\psi_{n+2}(u)}} \\ &= I(u) - (n+1)\cdot(n+1)!\,\cdot\,\psi_{n+2}(u) \end{align} Thus applies: \begin{align} I(u) &= (n+1)\cdot(n+1)!\,\cdot\,\psi_{n+2}(u) + (n+2)! \cdot \psi_{n+3}(u) \color{white}{\frac{1}{2}} \\ \\ I &= I(u) \ |_0^n = I(n) - I(0) \\ &= (n+1)! \cdot (n+1) \,\underbrace{\left[\psi_{n+2}(n) - \psi_{n+2}(0)\right]}_{\color{red}{=\,0; \text{ $n$ is even}}} \color{white}{\frac{1}{2}} \\ &\hspace{15pt} + (n+2)! \left[\psi_{n+3}(n) - \psi_{n+3}(0)\right] \color{white}{\frac{1}{2}} \\ &= (n+2)! \left[\psi_{n+3}(n) - \psi_{n+3}(0)\right] \end{align} But what is now \begin{align} \psi_{n+3}(n) \ ? \end{align}

Physics
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    The expansion of the Bernoulli polynomials of the second kind into a Newton series reads \begin{align} \psi_n(x) = \sum_{k=0}^n G_k \binom{x}{n-k} \end{align} We get: \begin{align} \psi_n(n-t) &= \sum_{k=0}^n G_k \binom{n-t}{n-k} \ &= \sum_{k=0}^{t-1} G_k \underbrace{\binom{n-t}{n-k}}{{=,0 \ \text{for k < t}}} + \sum{k=t}^n G_k \binom{n-t}{n-k} \ &= \sum_{k=0}^{n-t} G_k \binom{n-t}{n-t-k} \ &= \psi_{n-t}(n-t) \end{align} Thereby follows: \begin{align} \psi_n(n-3) &= \psi_{n-3}(n-3) = ? \end{align} Unfortunately, this does not get us anywhere... – Physics Nov 12 '21 at 15:02

1 Answers1

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According to Wikipedia: \begin{align} \psi_{2k}(k-1+y) = \psi_{2k}(k-1-y) \end{align} If we now choose $y = k - 2$, it follows: \begin{align} &\psi_{2k}(k-1+(k-2)) = \psi_{2k}(k-1-(k-2)) \color{white}{\frac{1}{2}}\\ & \Longrightarrow \hspace{10pt} \psi_{2k}(2k-3) = \psi_{2k}(1) \color{white}{\frac{1}{2}} \\ \end{align} If we consider an even $n$ as described in my problem, i.e. $n = 2k$, the following holds: \begin{align} &\psi_{n}(n-3) = \psi_{n}(1) = G_{n-1} + G_n \\ &\psi_{n+3}(n) = \psi_{n+3}(1) = G_{n+2} + G_{n+3} \end{align} We get for the integral: \begin{align} I &= (n+2)! \cdot \left[\psi_{n+3}(n) - \psi_{n+3}(0)\right] \color{white}{\frac{1}{2}}\\ &= (n+2)! \cdot \left[G_{n+2} + G_{n+3} - G_{n+3}\right] \color{white}{\frac{1}{2}} \\ &= (n+2)! \cdot G_{n+2} \end{align} Thus, for Newton-Cotes formulas: \begin{align*} E_n[f] = \left\{\begin{array}{ll} \displaystyle -f^{(n+1)}(\xi)\,h^{n+2}\cdot 2|G_{n+2}|, & n \text{ ungerade} \\ \displaystyle -f^{(n+2)}(\xi)\,h^{n+3}\cdot G_{n+2}, & n \text{ ungerade} \end{array}\right. \end{align*}

Physics
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