Let $\sigma >0$ be fixed. For even $k \in \mathbb{N} \cup \{0\}$, we consider the polynomial \begin{equation} \varphi_k(x) = \sum_{j=0}^{k} (-1)^j {k \choose j} b_j \, x^{2j} \quad x \in (-1,1), \end{equation} where \begin{equation} b_j = \frac{\Big(k+\sigma+\frac12\Big)_j}{\Big(\frac12 \Big)_j} \end{equation} and for $s \in \mathbb{R}$, $(s)_j$ denotes the Pochhammer symbol \begin{equation} (s)_{j}={\begin{cases}1&j=0\\s(s+1)\cdots (s+j-1)&j>0.\end{cases}} \end{equation} In particular, $\varphi_0(x) =1$.

My question is the following.

For $-1 < a < b < 1$, does there exist $c = c(a, b, \sigma)>0$ such that\begin{equation} \int_a^b \varphi_k(x)^2 dx \geq c \quad \end{equation}for any even $k \in \mathbb{N} \cup \{0\}$?

Unless I am mistaken, a straightforward computation yields \begin{equation} \int_a^b \varphi_k(x)^2 dx = \sum_{j=0}^k \sum_{\ell=0}^k (-1)^{j+\ell} {k \choose j} {k \choose \ell} \frac{b_j b_{\ell}}{2(j+\ell)+1} \, (b^{2(j+\ell)+1}-a^{2(j+\ell)+1}). \end{equation} But I do not see how I may bound this double sum from below.

**Remark 1:** I dont know if it is of any use, one may notice that $\varphi_k$ is a hypergeometric function of the form ${}_{2}F_{1}(-k,k+\sigma+\frac12;\frac12;x^2)$ (see https://en.wikipedia.org/wiki/Hypergeometric_function).

**Remark 2:** Using this interpretation as a hypergeometric function (which terminates), it is in fact possible to relate $\varphi_k$ to the *Jacobi polynomials* (https://en.wikipedia.org/wiki/Jacobi_polynomials):
\begin{equation}
\varphi_k(x) = {}_{2}F_{1}(-k,\sigma +\frac12 +k;\frac12; x^2)={\frac {k!}{(\alpha +1)_{k}}}P_{k}^{(-\frac12 ,\sigma )}(1-2x^2).
\end{equation}
Perhaps this observation may be of use.

*Comment*: *The same question is open for any odd $k \in \mathbb{N}$, but this time one considers*
\begin{equation}
\varphi_k(x) = \sum_{j=0}^{k} (-1)^j {k \choose j} c_j \, x^{2j+1},
\end{equation}
*with* $
c_j = \frac{\Big(k+\sigma+\frac32\Big)_j}{\Big(\frac32 \Big)_j} $.
*I suspect that the methodology is similar as for the case where $k$ is even.*