I have to solve the following exercise:
Suppose $X\sim N(\mu,1)$ and consider $Y=\dfrac{1-\Phi(X)}{\phi(X)}$, where $\phi,\Phi$ are the pdf and cdf of the standard normal distribution with mean zero. Compute the expected value of $Y$ in terms of $\mu$.
Please give me some hints only.
Let $Z$ be a standard normal variable. Since: $$\frac{1-\Phi(x)}{\phi(x)}=\frac{1}{\mathbb{E}[Z\,|\,Z>x]}=e^{x^2/2}\int_{x}^{+\infty}e^{-z^2/2}\,dz=\int_{0}^{+\infty}e^{-zx}e^{-z^2/2}\,dz$$ we have that: $$\mathbb{E}[Y]=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{+\infty}e^{-(x-\mu)^2/2}\int_{0}^{+\infty}e^{-zx}e^{-z^2/2}\,dz\,dx\tag{1} $$ so: $$\mathbb{E}[Y]=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{+\infty}\int_{0}^{+\infty}e^{-x^2/2}e^{-z^2/2}e^{-z(x+\mu)}\,dz\,dx \tag{2}$$ and switching the order of integration we get: $$\mathbb{E}[Y] = \int_{0}^{+\infty}e^{-z\mu}\,dz =\frac{1}{\mu}\tag{3}$$ as wanted.