I am trying to solve the following integral: $$\int_{-1}^{1}(1-x^2)^n e^{-\beta(x-\mu)^2}dx$$ where $x\in[-1,1]$, $\{\beta,\mu\}\in \mathbb{R}$, $n\in \mathbb{N}$ and $\beta >0$.
I tried to solve it through expanding $(1-x^2)^n=\sum_{k=0}^{n}(-1)^k\binom{n}{k}x^{2k}$ by binomial expantion then by solving the integrals with the form $$\int_{-1}^{1}x^{2k} e^{-\beta(x-\mu)^2}dx$$ where k=0, 1, 2, ... n, but i could not find the solution, in book "Table of Integrals, Series, and Products, EH II 195(31)" I could only found the solution for the integral with limits fron $-\infty $ to $\infty$ as $$\int_{-\infty}^{\infty}x^{n} e^{-(x-\mu)^2}dx=(2i)^{-n}H_n(i\mu),$$ which does not help. I tried to solve $\int_{-1}^{1}x^{2k} e^{-\beta(x-\mu)^2}dx$ by applying the integration by parts 2k times and i got the following solution $$\int_{-1}^{1}x^{2k} e^{-\beta(x-\mu)^2}dx=\frac{1}{\beta^{k+1}}\int_{-\beta}^{\beta}X^{2k} e^{-(X-\nu)^2}dX=-\frac{\sqrt{\pi}}{2\beta^{k+1}}\left[\sum_{j=0}^{2k}(-1)^jX^{2k^{(j)}}I^j(erfc(X-\nu))\right]_{-\beta}^{\beta}$$ where $X^{2k^{(j)}}$ is the $j^{th}$ derivative of $X^{2k}$ and $I^j(erfc(X-\nu))$ is the $j^{th}$ indefinite integral of $erfc(X-\nu)$, which is different that $i^n erfc(z)$ at DLMF 7.18.2, which is defined for integral from $z$ to $\infty$ hòw to get a form for $$I^j(erfc(X-\nu))=\left(\int\left(\int..._j...\left(\int erfc(X-\nu)dx \right)..._j...dx\right) dx\right) $$ I also checked the integral representation of the conflouent hypergeometric functions M(a,b,z) and U(a,b,z) but i could not reach a similar forms.
The exponential generating function of your sequence of integrals is
$$ \eqalign{\sum_{n=0}^\infty \frac{z^n}{n!} \int_{-1}^1 (1-x^2)^n e^{-\beta (x-\mu)^2}\; dx &= \int_{-1}^1 e^{z(1-x^2) -\beta (x-\mu)^2}\; dx\cr &= \sqrt{\frac{\pi}{4(\beta+z)}} e^{z - \beta z \mu^2/(\beta+z)} \left( \text{erf}\left(\frac{\beta \mu + \beta + z}{\sqrt{\beta+z}}\right)+ \text{erf}\left(\frac{-\beta \mu + \beta + z}{\sqrt{\beta+z}}\right)\right) }$$
EDIT: The term for $n$ is the $n$'th derivative of the exponential generating function evaluated at $z=0$. Unfortunately that's going to be a very complicated expression even for $n=10$. However, there are other things you can do with it. For example, you can get a linear recurrence relation:
$$ \beta^2 a(n+3) + (2 \beta \mu^2 - 2 \beta + 4 n + 9) \beta a(n+2) + (n+2)(n+1) (2 n - 4 \beta + 3) a(n+1) - 2 (n+2)(n+1) a(n) = 0 $$
where $a(n)$ is your integral.