# derivative of an integral coming from an expectation

I just wanted to check on the following result from this paper (a very rough sketch of the proof is given in the supplementary material, starting from eq 8).

$$\left.\frac{\partial}{\partial c} \left(\frac{1}{2 \pi q} \int \phi(\sqrt{q} z_1) \phi\left(\sqrt{q} u\right) e^\frac{-z_1^2}{2} e^\frac{-z_2^2}{2} \mathrm{d}z_1 \mathrm{d}z_2\right)\right|_{c=1} = \frac{1}{\sqrt{2 \pi}} \int \phi'(\sqrt{q} z)^2 e^{\frac{-z^2}{2}} \mathrm{d}z,$$

where $u = c z_1 + \sqrt{1 - c^2} z_2$.

So I started to write down the derivative

\begin{align} & = \frac{1}{2 \pi q} \int \phi(\sqrt{q} z_1) \frac{\partial}{\partial c} \left(\phi\left(\sqrt{q} u\right) \right) e^\frac{-z_1^2}{2} e^\frac{-z_2^2}{2} \mathrm{d}z_1 \mathrm{d}z_2 \\ & = \frac{1}{2 \pi q} \int \phi(\sqrt{q} z_1) \phi'\left(\sqrt{q} u\right) \sqrt{q} \frac{\partial u}{\partial c} e^\frac{-z_1^2}{2} e^\frac{-z_2^2}{2} \mathrm{d}z_1 \mathrm{d}z_2 \\ & = \frac{1}{2 \pi q} \int \phi(\sqrt{q} z_1) \phi'\left(\sqrt{q} u\right) \sqrt{q} \left(z_1 - \frac{c z_2}{\sqrt{1 - c^2}}\right)e^\frac{-z_1^2}{2} e^\frac{-z_2^2}{2} \mathrm{d}z_1 \mathrm{d}z_2 \\ & = \frac{1}{2 \pi q} \int \phi(\sqrt{q} z_1) \phi'\left(\sqrt{q} u\right) \sqrt{q} z_1 e^\frac{-z_1^2}{2} e^\frac{-z_2^2}{2} \mathrm{d}z_1 \mathrm{d}z_2 \\ & \qquad - \frac{1}{2 \pi q} \int \phi(\sqrt{q} z_1) \phi'\left(\sqrt{q} u\right) \sqrt{q} \frac{c z_2}{\sqrt{1 - c^2}} e^\frac{-z_1^2}{2} e^\frac{-z_2^2}{2} \mathrm{d}z_1 \mathrm{d}z_2 \\ & = \frac{1}{2 \pi} \int \phi'(\sqrt{q} z_1) \phi'\left(\sqrt{q} u\right) e^\frac{-z_1^2}{2} e^\frac{-z_2^2}{2} \mathrm{d}z_1 \mathrm{d}z_2 \\ & \qquad - \frac{1}{2 \pi q} \int \phi(\sqrt{q} z_1) \phi'\left(\sqrt{q} u\right) \sqrt{q} \frac{c z_2}{\sqrt{1 - c^2}} e^\frac{-z_1^2}{2} e^\frac{-z_2^2}{2} \mathrm{d}z_1 \mathrm{d}z_2, \end{align}

but then I got stuck with this second term. If we replace $c = 1$ in the expression, the expected result appears in the first term, but I have no idea how the second term should become zero. Definitely not with the factor $\frac{c}{\sqrt{1 - c^2}}$ in that second term.

Would anybody see where I made a mistake, missed something or how to proceed to get the desired result?

I have found my own mistake: I was a little to fast in using the identity $\int f(x) x e^{\frac{-x^2}{2}} = \int f'(x) e^{\frac{- x^2}{2}}$ in the last line. Acutally, the last line should have been
\begin{align} & = \frac{1}{2 \pi} \int \left[\phi'(\sqrt{q} z_1) \phi'\left(\sqrt{q} u\right) + c \phi(\sqrt{q} z_1) \phi''(\sqrt{q}u)\right] e^\frac{-z_1^2}{2} e^\frac{-z_2^2}{2} \mathrm{d}z_1 \mathrm{d}z_2 \\ & \qquad - \frac{1}{2 \pi q} \int \phi(\sqrt{q} z_1) \phi'\left(\sqrt{q} u\right) \sqrt{q} \frac{c z_2}{\sqrt{1 - c^2}} e^\frac{-z_1^2}{2} e^\frac{-z_2^2}{2} \mathrm{d}z_1 \mathrm{d}z_2, \end{align}
because $u$ also depends on $z_1$, which I did not account for in the last line of my question. To get to the final result, we need to apply the identity again for the last term
\begin{align} & = \frac{1}{2 \pi} \int \phi'(\sqrt{q} z_1) \phi'\left(\sqrt{q} u\right) e^\frac{-z_1^2}{2} e^\frac{-z_2^2}{2} \mathrm{d}z_1 \mathrm{d}z_2 \\ & \qquad + \frac{c}{2 \pi} \int \phi(\sqrt{q} z_1) \phi''(\sqrt{q}u) e^\frac{-z_1^2}{2} e^\frac{-z_2^2}{2} \mathrm{d}z_1 \mathrm{d}z_2 \\ & \qquad - \frac{c}{2 \pi} \int \phi(\sqrt{q} z_1) \phi''\left(\sqrt{q} u\right) e^\frac{-z_1^2}{2} e^\frac{-z_2^2}{2} \mathrm{d}z_1 \mathrm{d}z_2 \\ & = \frac{1}{2 \pi} \int \phi'(\sqrt{q} z_1) \phi'\left(\sqrt{q} u\right) e^\frac{-z_1^2}{2} e^\frac{-z_2^2}{2} \mathrm{d}z_1 \mathrm{d}z_2, \\ \end{align}