On the vanishing of the integral $\int^{n+4}_0 dt P_{\ell}\left(1 - \frac{2t}{n+4}\right)(-t+2)_{n+1}$ for $\ell \geq n+2$

Question

I came across the following integral \begin{equation} \int^{n+4}_0 dt P_{\ell}\left(1 - \frac{2t}{n+4}\right)(-t+2)_{n+1} \, , \end{equation} where in the above $P_{\ell}(x)$ is the Legendre polynomial, and $(x)_a$ is the Pochhammer symbol, $(x)_a = \tfrac{\Gamma(x+a)}{\Gamma(x)}$. $\ell$ is a non-negative integer number and $n$ is an integer bigger or equal to $-1$.

After running it for some values in Mathematica it is easy to observe that it vanishes when

• $n+\ell$ is any even number,
• $\ell \geq n+2$.

The command that defines the integral is

int[n_, l_] := 
 Integrate[
  LegendreP[l, 1 - (2 t)/(n + 4)] Pochhammer[-t + 2, n + 1], {t, 0, 
   n + 4}]

and then two examples for $n=-1$ and $n=0$

Table[int[-1, l], {l, 0, 5}] // FunctionExpand
Table[int[0, l], {l, 0, 5}] // FunctionExpand

and of course, more values can be checked in a straightforward manner.

I managed to prove that the integral yields $0$ when $n+\ell$ is equal to any even number, however, I have not managed to derive a proof that the integral vanishes for $\ell \geq n+2$.

I was wondering if there are any suggestions on how to prove that the integral, also, vanishes for $\ell \geq n+2$.

The proof I managed that shows the integral being equal to $0$ when $n + \ell$ is any even number is as follows:

• make the substitution \begin{equation*} t = -t^{\prime}+n+4 \, , \end{equation*}
• use the identity $P_{\ell}(-x) = (-1)^{\ell}P_{\ell}(x)$ in order to re-write \begin{equation} P_{\ell}\left(-1 + \frac{2t^{\prime}}{n+4}\right) = P_{\ell}\left(- \left(1 - \frac{2t^{\prime}}{n+4}\right) \right) = (-1)^{\ell} P_{\ell}\left(1 - \frac{2t^{\prime}}{n+4}\right) \end{equation}
• we can use $(-x)_m = (-1)^m (x-m+1)_m$ in order to re-express the Pochhammer as \begin{equation} (2+t^{\prime}-n-4)_{n+1} = (-2 + t^{\prime} -n)_{n+1} = (-1)^{n+1} (-t^{\prime}+2)_{n+1} \end{equation}

Having all the above, we can express the original integral in terms of the $t^{\prime}$ variable in the following manner: \begin{equation} \int^0_{n+4} - dt^{\prime} (-1)^{\ell} P_{\ell}\left(1 - \frac{2t^{\prime}}{n+4}\right) (-1)^{n+1} (-t^{\prime}+2)_{n+1} \end{equation} and we can now swap the integration limits in the above to get \begin{equation} (-1)^{1+n+\ell} \int^{n+4}_0 dt^{\prime} P_{\ell}\left(1 - \frac{2t^{\prime}}{n+4}\right) (-t^{\prime}+2)_{n+1} \, . \end{equation} In the above we rename $t^{\prime} \rightarrow t$ and get \begin{equation}\label{eq: integral_aux_02} (-1)^{1+n+\ell} \int^{n+4}_0 dt P_{\ell}\left(1 - \frac{2t}{n+4}\right)(-t+2)_{n+1} \, . \end{equation} So, what we have obtained so far is the following: \begin{equation} \begin{aligned} &\int^{n+4}_0 dt P_{\ell}\left(1 - \frac{2t}{n+4}\right)(-t+2)_{n+1} = \\ (-1)^{1+n+\ell} &\int^{n+4}_0 dt P_{\ell}\left(1 - \frac{2t}{n+4}\right) (-t+2)_{n+1} \Rightarrow \\ \left(1 - (-1)^{1+n+\ell} \right) &\int^{n+4}_0 dt P_{\ell}\left(1 - \frac{2t}{n+4}\right)(-t+2)_{n+1} = 0 \Rightarrow \\ \left(1 + (-1)^{n+\ell} \right) &\int^{n+4}_0 dt P_{\ell}\left(1 - \frac{2t}{n+4}\right)(-t+2)_{n+1} = 0 \, , \end{aligned} \end{equation} but for $n+\ell$ an even number $\left(1 + (-1)^{n+\ell} \right) =2$ and hence
\begin{equation} \int^{n+4}_0 dt P_{\ell}\left(1 - \frac{2t}{n+4}\right)(-t+2)_{n+1} = 0 \, . \end{equation}

Answer 1

It is true that if $\ell\ge n+2$, then $$\int_0^{n+4} dt P_{\ell}\left(1 - \frac{2t}{n+4}\right)(-t+2)_{n+1}=0$$

Proof :

Using $$P_{\ell}(x)=\sum_{k=0}^{\ell}\binom{\ell}{k}\binom{\ell+k}{k}\bigg(\frac{x-1}{2}\bigg)^k$$ (see here), we have

$$\begin{align}&\int_{0}^{n+4} dt P_{\ell}\left(1 - \frac{2t}{n+4}\right)(-t+2)_{n+1} \\\\&=\int_{0}^{n+4}dt\sum_{k=0}^{\ell}\binom{\ell}{k}\binom{\ell+k}{k}\bigg(-\frac{t}{n+4}\bigg)^k(-t+2)(-t+3)\cdots (-t+n+2) \\\\&=\sum_{k=0}^{\ell}\binom{\ell}{k}\binom{\ell+k}{k}\bigg(-\frac{1}{n+4}\bigg)^k\int_{0}^{n+4}t^k(-t+2)(-t+3)\cdots (-t+n+2)dt\tag1\end{align}$$ Let $a_j$ be the coefficient of $t^j$ in $(-t+2)(-t+3)\cdots (-t+n+2)$.

Then, $(1)$ can be written as $$\begin{align}(1)&=\sum_{k=0}^{\ell}\binom{\ell}{k}\binom{\ell+k}{k}\bigg(-\frac{1}{n+4}\bigg)^k\sum_{j=0}^{n+1}\frac{a_j(n+4)^{k+j+1}}{k+j+1} \\\\&=\sum_{k=0}^{\ell}\binom{\ell}{k}\binom{\ell+k}{k}(-1)^k\sum_{j=0}^{n+1}\frac{a_j(n+4)^{j+1}}{k+j+1} \\\\&=\sum_{j=0}^{n+1}a_j(n+4)^{j+1}\sum_{k=0}^{\ell}\binom{\ell}{k}\binom{\ell+k}{k}\frac{(-1)^k}{k+j+1} \\\\&=\sum_{j=0}^{n+1}a_j(n+4)^{j+1}G(\ell,j)\tag2\end{align}$$

where $$G(\ell,j):=\sum_{k=0}^{\ell}\binom{\ell}{k}\binom{\ell+k}{k}\frac{(-1)^k}{k+j+1}$$

WolframAlpha says $$G(\ell,j)=\frac{j!(\ell-j-1)!}{(-j-1)!(\ell+j+1)!}$$ which can be written as $$G(\ell,j)=\frac{\binom{\ell}{\ell+j+1}}{\ell \binom{\ell-1}{j}}$$

So, if $\ell\ge n+2$, then $G(\ell,j)=0$ for every $j=0,1,\cdots, n+1$.

Therefore, we have $(2)=0$.$\ \blacksquare$