What is the value of the following integral?
$$f(\alpha)=\int_0^\alpha [b-c \Phi^{-1}(1-\beta)] d\beta$$
where $0 \le \alpha\le 1$ , $b$ and $c$ are constants, and $\Phi$ is the cumulative distribution function (CDF) of the standard normal distribution.
Thanks.
The CDF of a normal function is given by
$$\Phi(u) = \frac12 \left [1+\operatorname*{erf}{\left (\frac{u}{\sqrt{2}} \right )} \right ] $$
So let $u = \Phi^{-1}(1-\beta)$; then $d\beta = -\Phi'(u) du = -\frac1{\sqrt{2 \pi}} e^{-u^2/2} du$. The integral then becomes
$$\frac1{\sqrt{2 \pi}}\int_{\Phi^{-1}(1-\alpha)}^{\infty} du \, (b-c u) e^{-u^2/2}$$
which I think you can take from here. I get that the integral is equal to
$$\alpha b - \frac{c}{\sqrt{2 \pi}} e^{-\left [\Phi^{-1}(1-\alpha) \right ]^2/2}$$