I am looking for a closed-form expression for the integral of the associated Legendre polynomial $P_l^m$ over the unit interval ($l \ge m$ non-negative integers), $$ I_l^m = \int_{0}^{1} P_l^m(x) \, dx \;. $$ There is an expression for $I_l^m$ in terms of Clausen's hypergeometric function $_3F_2$ (e.g. Gradshteyn & Ryzhik, 7.126.2), $$ I_l^m = \frac{(-1)^m \, \pi \, 2^{-2m-1}}{\Gamma\bigl(\frac{m+1}{2}\bigr) \, \Gamma\bigl(\frac{m+3}{2}\bigr)} \, \frac{(l+m)!}{(l-m)!} \, {_3F_2}\bigl(-\tfrac{l-m}{2}, \tfrac{m+l+1}{2}, \tfrac{m+2}{2}; m+1, \tfrac{m+3}{2}; 1\bigr) \;. \tag{$\star$} $$ However, when $l+m$ is even, $P_l^m$ is an even function, and $$ I_l^m = \frac{1}{2} \int_{-1}^{1} P_l^m(x) \, dx \;. $$ This integral has a well-known expression in terms of gamma functions, which yields the simple form $$ I_l^m = \frac{(-1)^m \, 2^{m-2} \, m \, \Gamma\bigl(\frac{l}{2}\bigr) \, \Gamma\bigl(\frac{l+m+1}{2}\bigr)}{\Gamma\bigl(\frac{l+3}{2}\bigr) \, \bigl(\frac{l-m}{2}\bigr)!} \qquad \text{($l+m$ even)} \;. \tag{$\dagger$} $$ The even case $(\dagger)$ follows from the hypergeometric function in $(\star)$ more or less directly using Saalschütz' theorem.
Is there a similarly simple expression for $I_l^m$ when $l+m$ is odd? The result on the Wolfram page seems incorrect, as it is merely $(\dagger)$ in a more complicated form, and I could not find it in any of the usual references. Looking at numerical results, I do not see a qualitative difference between the even and odd cases, so I still have some small hope for a positive answer.