# Is there a closed-form solution for the following integral?

The integral under consideration is as follows: $$F=\int_{a}^{1} \exp\Big\{c\Phi^{-1}(x+b) + d\Big\}\; \mathrm dx,$$ where $$0, and $$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.$$

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

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