The integral under consideration is as follows: $$ F=\int_{a}^{1} \exp\Big\{c\Phi^{-1}(x+b) + d\Big\}\; \mathrm dx, $$ where $0<a, b<1$, and $c>0, d\in\mathbb{R}$ are constants, and the notation $\Phi(\cdot)$ denotes the standard normal distribution function given by $$ \Phi(z) = \mathbb{P}(Z\leq z) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{z}e^{-\frac{u^2}{2}}\; \mathrm du. $$

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    $\begingroup$ Using the substitution $y=\Phi^{-1}(x+b)$, you should be able to work out the closed form solution. $\endgroup$
    – Gordon
    Jun 22, 2021 at 15:21
  • $\begingroup$ Thank you very much. It seems that it works. $\endgroup$
    – user53249
    Jun 22, 2021 at 16:51

1 Answer 1


Thanks to Gordon's help, we have that \begin{eqnarray*} F=exp\Big\{d + \frac{{c}^2}{2}\Big\}\Big[ \Phi\Big(\Phi^{-1}\Big(1+b\Big)-{c}\Big)- \Phi\Big(\Phi^{-1}\Big(a+b\Big)-{c}\Big)\Big] \end{eqnarray*}


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