I encounter the following integral, where $F^{-1}(u)$ is the quantile function for $\mathcal{N}(0,1)$ distribution: for $k \in\mathbb{N}$:
$$\int\limits_{1/2}^1 F^{-1}(u) u^k(1-u)^k\,\mathrm{d}u$$
Could this be simplified to some easier form as a function of $k$? Or if this could not be derived to an explicit form, as $k\rightarrow\infty$, the integrand becoming small, could we say something about its decreasing order in terms of $k$.
If $$F^{-1}(u)=\sqrt{2}\, \text{erf}^{-1}(2 u-1)$$ $$I(k)=\int\limits_{\frac12}^1 F^{-1}(u) u^k(1-u)^k\,du$$ does not show any analytical form.
Computing it and making a quick and dirty linear regression $$\log\big[I(k)\big]=a+ b\,k$$ for $0 \leq k \leq 50$ gives with $R^2=0.999919$ $$\begin{array}{clclclclc} \text{} & \text{Estimate} & \text{Standard Error} & \text{Confidence Interval} \\ a & -2.72809 & 0.11301 & \{-2.95532,-2.50086\} \\ b & -1.44222 & 0.00389 & \{-1.45005,-1.43439\} \\ \end{array}$$
Edit
One thing we can try is to let $u=\frac {x+1} 2$ to make $$I(k)=2^{-2 k-\frac{1}{2}}\int_0^1 \text{erf}^{-1}(x) \left(1-x^2\right)^k\,dx$$ and use the series expansion $$\text{erf}^{-1}(x)=\sum_{n=0}^\infty \frac{c_n}{2 n+1} \Big[\sqrt{\frac{\pi }{2}} x\Big]^{2n+1}$$ where $$c_0=1 \quad \text{and}\quad c_n=\sum _{m=0}^{n-1} \frac{c_m \,c_{n-m-1}}{(m+1) (2 m+1)}$$ $$I(k)=2^{-2 k-\frac{1}{2}}\sum_{n=0}^\infty \frac{2^{-n-\frac{3}{2}} \pi ^{n+\frac{1}{2}} \Gamma (2 k+1) \Gamma (n+1)}{(2 n+1) \Gamma (2 k+n+2)}c_n$$ Using a few terms, the asymptotics is $$I(k)=\frac{\sqrt{\pi } }{k}\,2^{-2 k-\frac{5}{2}}\exp\Big[-\frac{12-\pi }{12 k} \Big]$$
For $k=5$, this gives $I_5=0.00005280$ while numerical integration gives $0.00005326$.