For $t \in [-1,1]$, $\alpha = \arcsin(t)$, and integer $n \ge 0$, define $I_n(t)$ by $$ \begin{split} I_n(t) &= \frac{1}{\pi}\int_{-\pi}^\pi \cos^n\theta\sin^n(\theta + \alpha)d\theta = \frac{1}{\pi}\int_{-\pi}^\pi (\sin(2\theta+\alpha)+t)^nd\theta\\ &= \sum_{k = 0}^{\lfloor n/2 \rfloor} \sum_{j = 0}^{k } {n\choose 2k}{2k\choose 2j} S(2j, 2k - 2j) t^{n - 2k}t^{2j}(1 - t^2)^{k - j}. \frac{1}{2^{n + 2k -1}}, \end{split} $$ where $S(m,n) := \dfrac{(2m)!(2n)!}{m!n!(m+n)!}$ and the last equality in the above series of identities is thanks to this ME post https://math.stackexchange.com/a/4242413/168758.
Question. What is are analytic formula for $I_n(0)$, $I_n(1)$, and $I_n'(0)$, perhaps in terms of special functions ?
Observations
- If $n=0$, then $I_n(t) \equiv 2\pi$, and so $I_n(0) = I_n(1) = 2$ and $I_n'(0) = 0$.
- If $n=1$, then $I_n(t) \equiv t$, and so $I_n(0) = 0$, and $I_n(1) = I_n'(0) = 1$
- If $n=2$, then $I_2(t) \equiv (2t^2 + 1)/4$, and so $I_n(0) = 1/4$, $I_n(1) = 3/4$ and $I_n'(0) = 0$,
- If $n=3$, then $I_n(t) \equiv t(2t^2 + 3)/8$, and so $I_n(0) = 0$, $I_n(1) = 5/8$, and $I_n'(0) = 3/8$.
- ...
There is a one-line comment by user Maxim, under the original question here https://math.stackexchange.com/a/4242413/168758, namely that $$ \begin{split} I_n(t) = 2^{1-n}(-i\cos\alpha)^n P_n(i\tan\alpha)=2^{1-n}(-i)^n(1-t^2)^{n/2}P_n(it(1-t^2)^{-1/2}), \end{split} $$ where $P_n$ is the $n$th Legendre polynomial and $i := \sqrt{-1}$. It follows directly that $I_n$ is $\mathcal C^\infty$ on $(-1,1)$ with derivative $$ \begin{split} I_n'(t) = &n2^{1-n}(-i)^{n+1}t(1-t^2)^{n/2-1}P_n(it(1-t^2)^{-1/2})\\ &\quad + 2^{1-n}(-i)^n(1-t^2)^{n/2}(i(1-t^2)^{-1/2}+it^2(1-t^2)^{-3/2})P_n'(it(1-t^2)^{-1/2}), \end{split} $$ for all $t \in (-1,1)$. In particular, one computes $$ \begin{split} I_n(0) &= 2^{1-n}(-i)^n P_n(0),\\ I_n(1) &= 2^{1-n}P_n(0)\delta_{n=0},\\ I_n'(0) &= 2^{1-n}(-i)^{n+1} P_n'(0). \end{split} $$
As sanity check, noting that $P_2(x) = (3x^2-1)/2$, we have that for $n=2$, $$ \begin{split} I_2(t) &= 2^{1-2}(-i)^2(1-t^2) P_2(-it(1-t^2)^{-1/2})\\ &=-2^{-1}(1-t^2)(3(-it(1-t^2)^{-1/2})^2-1)/2\\ &= 2^{-1}(3t^2 + 1-t^2)/2 = (2t^2+1)/4, \end{split} $$ which corresponds to the formula stated in the question.