Write [tex]L= \frac{1}{2} (\frac{x-x'}{\lambda})^{2} - V(x)[/tex] , change integration variable to [tex]x-x'=y[/tex] , then expand [tex]V(x),\phi(x)[/tex] about [tex]y=0[/tex] and keep first order terms in [tex]\lambda[/tex] ;
[tex]\phi(x',t+\lambda)= A \int dy exp(\frac{iy^{2}}{2\lambda}) exp[-i\lambda (V(x') + yV'(x') + ..)][\phi(x',t) + y \frac{\partial {\phi(x')}}{\partial{x'}} + \frac{1}{2} y^{2}\frac{\partial^{2} \phi(x',t)}{\partial {x'}^{2}} +...][/tex]
or
[tex]\phi(x',t+\lambda) = A[1 - i \lambda V(x')] \int dy exp[\frac{iy^{2}}{2\lambda}][\phi(x',t) + (1/2) y^{2} \frac{\partial^{2}\phi}{\partial {x'}^{2}}][/tex]
we dropped the factor
[tex]exp[-iV'(x')y\lambda][/tex]
because when the integration over y is done this will be of order
[tex]\lambda^{3/2}[/tex]
we also used
[tex] \int dy y exp(-a y^{2}) = 0[/tex]

Now use the integrals:
[tex] \int dy exp(-a y^{2}) = (\pi/a)^{1/2}[/tex]
[tex] \int dy y^{2} exp(-a y^{2}) = (\pi/2)^{1/2}(1/2a)[/tex]

Schrodinger eq follows when you choose A so that the coefficient of
[itex]\phi(x',t)[/itex] is one, and take the limit [itex]\lambda\rightarrow 0[/itex];

Thanks for the reply!
I didn't know the bounds are infinities... Also, I couldn't get [tex]\lambda[/tex] to the 3/2th power, for [tex]-iV'(x') y \lambda[/tex], but as you did, I used only the first term for potential.

Anyway, I couldn't see how Schrödinger equation comes out from what you've written, so
I've tried to work out the details for myself, but I ended up a weird equation that is far from Schrödinger's. Here it goes:

(You seem to've cranked up [tex]\hbar[/tex] and m to 1, but for keeping a better track of things, I'm leaving them there).

[tex]L = \frac{m}{2} {(\frac{x-x'}{\epsilon})}^2 - V(x)[/tex]
[tex]y = x-x'[/tex], keeping [tex]x'[/tex] fixed, [tex]dy = dx[/tex] (I guess this was what you meant).

[tex]A \sqrt{\frac{\pi}{-im/\hbar \epsilon}} = 1[/tex]. [tex]V(x') \frac{\partial^2 \varphi}{\partial x^2}...[/tex] term is dropped since it has an second order [tex]\epsilon[/tex]. OK, by doing so, I had the potential term nicely, but I still have a weird extra coefficent in the same place. This time, it's this:

the error was in
[tex] \int dy e^{-a y^{2}} = (\pi/2)^{1/2}(1/2a)[/tex]
it should be
[tex] \int dy e^{-a y^{2}} = (\pi/a)^{1/2}(1/2a)[/tex]
I think my miss-keying is to blame

Horray! Huge thanks, finally Schrödinger equation did come out!
One last question, how could you (or Feynman) see which term you had to expand? I rather tried expanding [tex]e^{iLdt/\hbar}[/tex] directly, without expanding the potential term around a fixed point, and without leaving a [tex]e^{imv^2/2\hbar}[/tex]. It was all a mess. I really wonder how you did see?