Feynman's derivation of Schrödinger equation. Potential spatial dependence I am working on the book "Quantum Mechanics and Path Integrals" from Feynman and Hibbs. When finding the correspondence with Schrödinger equation he takes 

$$\eqalign{&\psi(x,t+\epsilon) = {}\cr
&\int_{-\infty}^{\infty} 
\!\!\exp\left\{\frac{i\,\epsilon}{\hbar} 
L\left(\frac{x + y}{2},\frac{x - y} {\epsilon}\right) \right\}\,
\psi(y,t)\,\frac{\mathrm{d}y}{A(\epsilon)}\cr}$$

Making the Lagrangian explicitly as $L = m\dot{x}^2/2 - V(x,t)$, and making the substitution $y = x + \eta$  he gives 

$$\eqalign{
&\psi(x,t+\epsilon) = \int_{-\infty}^{\infty}
\exp\left\{ \frac{i m \eta^2}{2\hbar\epsilon} \right\} \cr
&\qquad\exp\left\{ -\frac{i\, \epsilon}{\hbar}  V\left( x+ \frac{\eta}{2}, t \right)   \right\}     \psi(x +\eta, t) \, \frac{\mathrm{d}\eta}{A(\epsilon)}\cr}$$

Now the first exponential varies very rapidly and he says that most of the integral will be contributed by $\eta$ in the order of 0 to $\sqrt{2\hbar \epsilon/m}$. For a small $\eta$ he can now expand the second exponential, as well as $\psi$

$$\eqalign{
&\psi(x,t) + \epsilon\, \frac{\partial \psi}{\partial t} = {}\cr
&\quad\int_{-\infty}^{\infty} 
\exp\left\{ \frac{i m \eta^2}{2\hbar\epsilon} \right\}   \left[1 -\frac{i\, \epsilon}{\hbar} V \left( x, t \right)\right] \cr 
&\qquad\left[\psi(x,t) + \eta \frac{\partial \psi}{\partial x}  +\frac{\eta^2}{2} \frac{\partial^2 \psi}{\partial x^2}   \right] \, \frac{\mathrm{d}\eta}{A(\epsilon)}\cr}$$

Here, he replaces $\epsilon V(x +\frac{\eta}{2},t)$ for $\epsilon V(x,t)$ saying that the error is of higher order than $\epsilon$.
My problem is that the expansion of $V(x +\frac{\eta}{2},t)$ would have a term of order $\eta$, which when multiplied by $\eta \frac{\partial \psi}{\partial x}$ would give a term of order $\eta^2$ and it's integration would be non-zero. The terms of order $\eta^2$ are not neglected, since that going with the second derivative of $\psi$ is preserved. The problematic term is then
$$\int_{-\infty}^{\infty} \exp\left\{ \frac{i m \eta^2}{2\hbar\epsilon} \right\}   \frac{i\, \epsilon \, \eta^2}{\hbar}  \left. \frac{\partial V}{\partial (x + \eta/2)} \right|_{(x,t)}   \frac{\partial \psi}{\partial x}  \, \frac{\mathrm{d}\eta}{A(\epsilon)}$$
I think that the problem might be I am not working properly the Taylor series.
Thank you for your help.
 A: Okay, the problem actually is not there. Both statements are correct, the mistake I've made was in the comparison between orders of the development.
We take just the first order in $\epsilon$ in the left hand side 
$$\psi + \epsilon \partial_t\psi,$$
and the last integral,
$$\int_{-\infty}^{\infty} \exp\left\{ \frac{i m \eta^2}{2\hbar\epsilon} \right\}   \frac{i\, \epsilon \, \eta^2}{\hbar}  \left. \frac{\partial V}{\partial (x + \eta/2)} \right|_{(x,t)}   \frac{\partial \psi}{\partial x}  \, \frac{\mathrm{d}\eta}{A(\epsilon)},$$
 will give something of the order of $\epsilon^2$, since we have 
$$\int_{-\infty}^{\infty} \exp\left\{ \frac{i m \eta^2}{2\hbar\epsilon} \right\}   \eta^2 \, \frac{\mathrm{d}\eta}{A(\epsilon)} = \frac{i\hbar \epsilon}{m},$$
where the condition for $A$ is found through the correspondence of the terms of zeroth order, and nothing else depends on $\eta$, and there is one factor $\epsilon$ already present.
The term with the second derivative does have an $\eta^2$, but only its product with the 1 in the expansion of the potential is preserved. 
The identification of the first order terms on $\epsilon$ gives the expression of the Schrödinger equation.
