Let f(x) be 1/6 for $0\leq x \leq 6$ and 0 elsewhere. Let g(x) be $x^2-3ix$
Find h(4) where h is the convolution of f and g.
Solution: $4-3i, h(x)=\frac{1}{6}(72-36x+6x^2+54i-18ix)=12+x^2+9i-6x-3ix$
So I tried to use the convolution theorem $f*g=L^{-1}(L(f)L(g))$.
I took the Laplace transform of f(x) between 0 and 6 and got
$$\frac{1}{6}(\frac{-e^{-6s}}{s}+\frac{1}{s})$$.
For g(x), I got $$\frac{2}{s^3}-\frac{3i}{s^2}$$.
The product of L(f) and L(g) gives $$\frac{1}{6}(\frac{-2e^{-6s}}{s^4}+\frac{2}{s^4}+\frac{3ie^{-6s}}{s^3}+\frac{-3i}{s^3})$$
Taking the inverse Laplace transform, I get
$$\frac{1}{6}(\frac{-(x-6)^3}{3}u(x-6)+\frac{x^3}{3}+\frac{3i(x-6)^2}{2}u(x-6)-\frac{3ix^2}{2})$$
(I applied the rule $L^{-1}(G(s)e^{-as})=g(t-a)u(t-a)$)
Evaluating at 4, u(-2) gives 0, and I'm left with $\frac{1}{6}(\frac{64}{3}-24i)$, which isn't the same as in the solution.
I also tried evaluating that convolution with the method explained here:
http://www.dtic.mil/dtic/tr/fulltext/u2/a602143.pdf
But the problem is that g(x) has no limit, i.e. the limits are negative and positive infinity, which doesn't seem to work.
Using the conventional way, i.e. integral from negative to positive infinity of $f(\tau)g(t-\tau)d\tau$, the problem is that g(x) isn't really a "nice" function. The convolution is symmetrical, but even so, I find it hard to find the limits of integrations because of the function g(x).
What did I do wrong ?
Thanks for your help !