I have this:
$(f*g)(x)=\int_{x-\frac12}^{x+\frac12} f(t)dt$
where $t = x-y$
and $f(x)$=\begin{cases}
0&\text{if}\, x\ < a\\
\exp(-x)&\text{if}\, a \leq x \
\
\end{cases}
I have problem with splitting this integral, the end result is supposed to be a function with three pieces. Any ideas? What I have trouble understanding is how to split this integral when we have $x$ in the interval.
Kind regards Pontus
Best Answer
Based on the information provided:
If $\> x+\frac{1}{2} < a \quad$ (i.e. $\> x < a-\frac{1}{2}$) \begin{align*} \int_{x-\frac{1}{2}}^{x+\frac{1}{2}} f(t) \mathrm{d} t &= \int_{x-\frac{1}{2}}^{x+\frac{1}{2}} 0 \> \mathrm{d} t = 0 \end{align*}
If $\> x-\frac{1}{2} > a \quad$ (i.e. $\> x > a +\frac{1}{2}$) \begin{align*} \int_{x-\frac{1}{2}}^{x+\frac{1}{2}} f(t) \mathrm{d} t &= \int_{x-\frac{1}{2}}^{x+\frac{1}{2}} \mathrm{e}^{-t} \mathrm{d} t = \mathrm{e}^{-(x-\frac{1}{2})} - \mathrm{e}^{-(x+\frac{1}{2})} \end{align*}
If $\> x \in \left( a-\frac{1}{2}, a+\frac{1}{2} \right) $ \begin{align*} \int_{x-\frac{1}{2}}^{x+\frac{1}{2}} f(t) \mathrm{d} t &= \int_{x-\frac{1}{2}}^{a} f(t) \mathrm{d} t + \int_{a}^{x+\frac{1}{2}} f(t) \mathrm{d} t = \int_{x-\frac{1}{2}}^{a} 0 \> \mathrm{d} t + \int_{a}^{x+\frac{1}{2}} \mathrm{e}^{-t} \mathrm{d} t = \mathrm{e}^{-a} - \mathrm{e}^{-(x+\frac{1}{2})} \end{align*}
Combine to get the resulting function with three pieces.