How do I evaluate $\displaystyle \int_{-\infty}^z e^{\frac{-t^2+2t\alpha\mu}{2\sigma^2\alpha^2}+\frac{\lambda t}{1-\lambda}} dt$ ??

Question

How do you evaluate: $$\displaystyle \int_{-\infty}^z e^{\frac{-t^2+2t\alpha\mu}{2\sigma^2\alpha^2}+\frac{\lambda t}{1-\alpha}} dt = ??$$ Many thanks.

Answer 1

$$ \displaystyle \int_{-\infty}^z e^{\frac{-t^2+2t\alpha\mu}{2\sigma^2\alpha^2}+\frac{\lambda t}{1-\alpha}} dt =e^{A B^2}\int_{-\infty}^z e^{-A (t-B)^2}dt\ , $$ with $A=1/(2\sigma^2\alpha^2)$ and $2AB=\frac{2\alpha \mu}{2\sigma^2\alpha^2}+\frac{\lambda}{1-\alpha}$. Solve for $B$. Next, make the change of variable $\sqrt{A}(t-B)=y$, yielding $$ \frac{e^{A B^2}}{\sqrt{A}}\int_{-\infty}^{\sqrt{A}(z-B)} e^{-y^2}dy=\frac{e^{A B^2}}{\sqrt{A}}\left(\int_{-\infty}^{+\infty} e^{-y^2}dy-\int_{\sqrt{A}(z-B)}^{+\infty} e^{-y^2}dy\right)\ . $$ Then, use the definition of the Complementary Error function https://en.wikipedia.org/wiki/Error_function as well as the Gaussian integral to express the two integrals above.