I need to calculate $(f*f)(x)$ of $f(x) = 1_{[0,1]}(x)$, which is the indicator function defined with
Calculating the integral $(f*f)(x) = \int_{0,}^{x}1_{[0,1]}(t) \cdot1_{[0,1]}(x-t) dt$ gives $(f*f)(x) = \begin{Bmatrix} x & \mathbb{for }\space x\in[0,1] \\ 0 & \mathbb{otherwise} \end{Bmatrix}$
but the Laplace transformation of $f(x)$ gives the convolution as $(f*f)(x) = x - 2 (x-1)\theta(x-1) + (x-2)\theta(x-2)$, where $\theta(x)$ is the Heavyside theta function.
What am I doing wrong?
Looking carefully at the definition of the integrand (for $x\in\mathbb{R}$), you can see that $f\ast f$ is $0$ for $x\notin[0,2]$, and that for $x\in[0,2]$ $$ f\ast f(x) = \int_{\max(0,x-1)}^{\min(1,x)} dt = \begin{cases} \int_{0}^{x} dt = x & x\leq 1 \\ \int_{x-1}^{1} dt = 2-x & x\geq 1 \end{cases} $$ As mentioned in a comment above, you can alternatively get this by seeing that you are computing the probability density function of $D=\mathcal{U}_{[0,1]}\ast\mathcal{U}_{[0,1]}$, where $\mathcal{U}_{[0,1]}$ is the uniform distribution on $[0,1]$. This is the distribution of a random variable $Y=X_1+X_2$, where the $X_i$'s are independent uniform r.v.s on $[0,1]$: in particular, $Y$ take values in $[0,2]$, and it is "easy" to see that its density is symmetric around $1$ and that the probability to be around $1/2$ is bigger than the probability to be around $0$ or $1$. (on a side note, $Y$ follows a triangular distribution on $[0,2]$ with mode $1$).