analytically calculate value of convolution at certain point i'm a computer science student and i'm trying to analytically find the value of the convolution between an ideal step-edge and either a gaussian function or a first order derivative of a gaussian function.
In other words, given an ideal step edge with amplitude $A$ and offset $B$:
$$
i(t)=\left\{ 
  \begin{array}{l l}
    A+B & \quad \text{if $t \ge t_{0}$}\\
    B & \quad \text{if $t \lt t_{0}$}\\
  \end{array} \right.
$$
and the gaussian function and it's first order derivative
$$
g(t) = \frac{1}{\sigma \sqrt{2\pi}}e^{- \frac{(t - \mu)^2}{2 \sigma^2}}\\
g'(t) = -\frac{t-\mu}{\sigma^3 \sqrt{2\pi}}e^{- \frac{(t - \mu)^2}{2 \sigma^2}}
$$
i'd like to calculate the value of both
$$
o(t) = i(t) \star g(t)\\
o'(t) = i(t) \star g'(t)
$$
at time $t_{0}$ ( i.e. $o(t_{0})$ and $o'(t_{0}) )$.
I tried to solve the convolution integral but unfortunately i'm not so matematically skilled to do it. Can you help me?
Thank you in advance very much.

We can write $i(t):=B+A\chi_{[t_0,+\infty)}$. Using the properties of convolution, we can see that 
\begin{align}
o(t_0)&=B+A\int_{-\infty}^{+\infty}g(t)\chi_{[t_0,+\infty)}(t_0-t)dt\\
&=B+A\int_{-\infty}^0\frac 1{\sigma\sqrt{2\pi}}\exp\left(-\frac{(x-\mu)^2}{2\sigma^2}\right)dx.
\end{align}
The integral can be expressed with erf-function. 
For the second one, things are easier since we can compute the integrals:
\begin{align}
o'(t_0)&=A\int_{-\infty}^0g'(t)dt=Ag(0).
\end{align}
