Consider the following function $$ f(\lambda) = \alpha (1+\lambda^2) + (1-\alpha)2\int_\lambda^\infty (x-\lambda)^2 \phi(x) dx $$ where $\alpha \in (0,1)$ and $\phi$ is the standard normal probability density function.
Is there a good (closed-form) approximation to the minimum value and/or minimizer of $f$ over $[0,\infty)$?
Using Leibniz's rule to differentiate the integral we have
$$\frac {\partial}{\partial \lambda} \int_\lambda^\infty (x-\lambda)^2 \phi(x) dx = -(\lambda-\lambda)^2 \phi(\lambda) + \int_\lambda^\infty \frac {\partial}{\partial \lambda}(x-\lambda)^2 \phi(x) dx$$ $$=-2\int_\lambda^\infty (x-\lambda)\phi(x) dx = -2\int_\lambda^\infty x\phi(x) dx +2\lambda\int_\lambda^\infty \phi(x) dx$$ The integrals can be transformed into truncated standard normals in $[\lambda, \infty]$
$$\frac {\partial}{\partial \lambda} \int_\lambda^\infty (x-\lambda)^2 \phi(x) dx =-2\Phi(-\lambda)\int_\lambda^\infty x\phi(x)\left[\Phi(-\lambda)\right]^{-1} dx +2\lambda\Phi(-\lambda)\int_\lambda^\infty \phi(x)\left[\Phi(-\lambda)\right]^{-1} dx$$
$$\Rightarrow \frac {\partial}{\partial \lambda} \int_\lambda^\infty (x-\lambda)^2 \phi(x) dx =-2\phi(\lambda) +2\lambda\Phi(-\lambda)\cdot 1$$
Then (simplyfying $2$) $$\frac {\partial f(\lambda)}{\partial \lambda} = \alpha\lambda - 2(1-\alpha)\left[\phi(\lambda) -\lambda\Phi(-\lambda)\right] = 0$$
$$\alpha\lambda = 2(1-\alpha)\left[\phi(\lambda) -\lambda\Phi(-\lambda)\right] \qquad [1]$$
From $[1]$ it is obvious that $\lambda^* \neq0$. Then
$$\frac {\alpha}{2(1-\alpha)}=\left[\phi(\lambda)\lambda^{-1} -\Phi(-\lambda)\right] = \phi(\lambda)\left[\lambda^{-1} -m(\lambda)\right]$$
where $m(\lambda)$ is Mills ratio of the standard normal. The traditional approximation to the mills ratio of the standard normal is $m(\lambda) = \lambda^{-1} - \lambda^{-3} +3\lambda^{-5} -...$ (for lots of stuff on Mills ratio and other approximation see here). Use just the first two terms to obtain
$$\frac {\alpha}{2(1-\alpha)}\approx \lambda^{-3}\phi(\lambda) = \lambda^{-3}\frac{1}{\sqrt{2\pi}}e^{-\frac 12 \lambda^2}$$
$$\Rightarrow \ln\left(\frac {\alpha}{2(1-\alpha)}\right) +\frac 12 \ln {2\pi}\approx -3\ln \lambda -\frac 12 \lambda^2$$
Approximate (at your peril) $\ln \lambda \approx \lambda -1$ to obtain
$$\ln\left(\frac {\alpha}{2(1-\alpha)}\right) +\frac 12 \ln {2\pi}\approx -3(\lambda -1)-\frac 12 \lambda^2$$
$$\Rightarrow \frac 12 \lambda^2 +3\lambda -3+\ln\left(\frac {\alpha}{2(1-\alpha)}\right) +\frac 12 \ln {2\pi} =0$$
which is a quadratic in $\lambda$ and the roots will be a function of $\alpha$.