Ok guys, I have a very trivial (not for me, though) question regarding the folowing statement:
$u_o (x) = \exp \left ( -\int^{x}_1 \frac{M \delta(y)}{2 \epsilon} dy \right)$
... I don't understand how the author gets to the following answers:
$u_0 = 1 $ if $x < 0$
or
$u_0 = \exp \left ( \frac{M}{2 \epsilon}\right)$ if $x > 0$
The Dirac delta impulse satisfies
$$\int_a^b\delta(x)dx=\begin{cases}1,&\text{if }[a,b]\text{ contains }0\\0,&\text{otherwise}\end{cases},\quad a<b$$
So in your case if $x<0$, then the integration interval contains the value $0$ and the integral becomes
$$\frac{M}{2\epsilon}\int_x^1\delta(y)dy=\frac{M}{2\epsilon}$$
Note that you get this value for $x<0$ and not for $x>0$, as suggested in your question. For $x>0$ the integral is $0$ and consequently $u_0(x)=1$.