Let $\Phi$ be standard Gaussian CDF and $u > 0$. What is good u-bound for $\int_0^1\Phi(u/r - ur)dr$ as a function of $u$?

Question

Let the function $\Phi(x) := (1/\sqrt{2\pi})\int_{-\infty}^x e^{-t^2/2}dt$ be the standard Gaussian CDF. For $u > 0$, define $I(u) := \int_0^1 \Phi(u/r-ur)dr$.

Question. What are good upper-bounds for $I(u)$ in terms of $u$ ?

Answer 1

A good simple upper bound for $u > 1$:

Using integration by parts and then the substitution $w = \frac{u}{r} - ur$, we have \begin{align} I(u) &= \frac{1}{2} + \frac{1}{2u\sqrt{2\pi}} \int_0^\infty \mathrm{e}^{-\frac{1}{2}w^2} (\sqrt{w^2+4u^2} - w)\mathrm{d} w\\ &\le \frac{1}{2} + \frac{1}{2u\sqrt{2\pi}} \int_0^\infty \mathrm{e}^{-\frac{1}{2}w^2} \left(2u + \frac{w^2}{4u} - w\right)\mathrm{d} w\\ &= 1 - \frac{1}{2u\sqrt{2\pi}} + \frac{1}{16u^2}. \end{align} The upper bound $I_1(u) = 1 - \frac{1}{2u\sqrt{2\pi}} + \frac{1}{16u^2}$ is good if $u > 1$. For example, $\frac{I_1(1)-I(1)}{I(1)} \approx 0.009176113058$, $\frac{I_1(2)-I(2)}{I(2)} \approx 0.0007002868670$, and $\frac{I_1(10)-I(10)}{I(10)} \approx 0.000001187547307$.