Is this estimate true or not true? Let $\varepsilon>0$. Let $\varphi(x)=\frac{1}{\sqrt{2\pi}}e^{-x^2/2}$ the standard normal density function. Then 
$$\lim_{\varepsilon\to 0}\int_0^1 \frac{1}{\sqrt{x}}\left[ \varphi\left(\frac{\sqrt{x}-\varepsilon}{\sqrt{x}}\right)-\varphi\left(\frac{\sqrt{x}+\varepsilon}{\sqrt{x}}\right)\right]dx=0.$$
The question I have is about the modulus of continuity. I would like to know if
$$\left|\int_0^1 \frac{1}{\sqrt{x}}\left[ \varphi\left(\frac{\sqrt{x}-\varepsilon}{\sqrt{x}}\right)-\varphi\left(\frac{\sqrt{x}+\varepsilon}{\sqrt{x}}\right)\right]dx\right|\leq C\varepsilon,$$
for some universal constant $C>0$.
Observe that pulling the absolute value inside the integral and using the Lipschitz property on $\varphi$ is too much, since $\frac{1}{x}$ is not integrable.
Thanks!!
 A: We can find the asymptotic behaviour via the substitution $t = 1/\sqrt{x}$. That substitution directly yields
$$\int_0^1 \frac{1}{\sqrt{x}} \biggl[\varphi\biggl(1 - \frac{\varepsilon}{\sqrt{x}}\biggr) - \varphi\biggl( 1+ \frac{\varepsilon}{\sqrt{x}}\biggr)\biggr]\,dx = 2\int_1^{+\infty} \frac{\varphi(1 - \varepsilon t) - \varphi(1 + \varepsilon t) }{t^2}\,dt.\tag{1}$$
Integration by parts on the right hand side then yields
\begin{align}
\int_1^{+\infty} \frac{\varphi(1 - \varepsilon t) - \varphi(1 + \varepsilon t)}{t^2}\,dt
&= \biggl[\frac{\varphi(1 + \varepsilon t) - \varphi(1 - \varepsilon t)}{t}\biggr]_1^{+\infty}\\
&\qquad\qquad - \varepsilon \int_1^{+\infty} \frac{\varphi'(1 + \varepsilon t) + \varphi'(1 - \varepsilon t)}{t}\,dt\\
&= \varphi(1 - \varepsilon) - \varphi(1 + \varepsilon)\\
&\qquad \qquad + \varepsilon \int_1^{+\infty} \frac{(1 + \varepsilon t)\varphi(1 + \varepsilon t) + (1 - \varepsilon t)\varphi(1 - \varepsilon t)}{t}\,dt\\
&= \varphi(1 - \varepsilon) - \varphi(1 + \varepsilon) + \varepsilon^2 \int_1^{+\infty} \varphi(1 + \varepsilon t) - \varphi(1 - \varepsilon t)\,dt\\
&\qquad\qquad + \varepsilon \int_1^{+\infty} \frac{\varphi(1+\varepsilon t) + \varphi(1 - \varepsilon t)}{t}\,dt\\
&= \varphi(1 - \varepsilon) - \varphi(1 + \varepsilon) + \varepsilon \int_{\varepsilon}^{+\infty} \varphi(1+u) - \varphi(1-u)\,du\\
&\qquad \qquad + \varepsilon \int_{\varepsilon}^{+\infty} \frac{\varphi(1+u) + \varphi(1-u)}{u}\,du,\tag{2}
\end{align}
where at the end we substituted $u = \varepsilon t$ in both integrals.
Clearly $\varphi(1-\varepsilon) - \varphi(1 + \varepsilon) = -2\varphi'(1)\cdot \varepsilon + O(\varepsilon^2)$, and
$$\int_{\varepsilon}^{+\infty} \varphi(1+u) - \varphi(1-u)\,du = \Phi(\varepsilon - 1) - \Phi(1 + \varepsilon) = \Phi(-1) - \Phi(1) + O(\varepsilon).$$
We write the last integral in $(2)$ as
$$ \int_1^{+\infty} \frac{\varphi(1+u) + \varphi(1-u)}{u}\,du + \int_{\varepsilon}^1 \frac{\bigl(\varphi(1+u) - \varphi(1)\bigr) + \bigl(\varphi(1-u)-\varphi(1)\bigr)}{u}\,du + 2\varphi(1)\int_{\varepsilon}^1\frac{du}{u}$$
and see that its value is
$$A + 2\varphi(1) \lvert \log \varepsilon\rvert + O(\varepsilon)$$
with
$$A = \int_1^{+\infty} \frac{\varphi(1+u) + \varphi(1-u)}{u}\,du + \int_{0}^1 \frac{\bigl(\varphi(1+u) - \varphi(1)\bigr) + \bigl(\varphi(1-u)-\varphi(1)\bigr)}{u}\,du.$$
Altogether, we find the asymptotic behaviour
$$4 \varphi(1) \varepsilon \lvert \log \varepsilon\rvert + 2\bigl(A -2\varphi'(1) + \Phi(-1) - \Phi(1)\bigr)\cdot \varepsilon + O(\varepsilon^2),$$
which isn't quite as good as you hoped for.
