I'm trying to find the result of the sum of two PDF that is the square of two continuous uniform distribution (rectangular distribution) that have $\mu =0$ and $\sigma =1$. The square of this continuous uniform distribution is defined from 0 to 3 and is $$p(x)= \frac{1}{2 \sqrt{3x}}$$ Now i need to find the PDF of the sum of two random variables with p(x) as distribution. The convolution integral is $$p(u)=\int_0^3 p(u-t)p(t)dt=\int_0^3 \frac{dt}{12 \sqrt{t(u-t)}}= \frac{arcsin \left( \sqrt{ \frac{3}{u}} \right)}{6}$$ This funcion is not defined on R from 0 to 6. If i extract the real part of $(p(u))$ i obtain the correct PDF from 0 to 3 but the part from 3 to 6 is wrong and p(u) is not normalized from 0 to 6. I simulate this PDF with python, here is an image
Simulation of PDF: https://i.stack.imgur.com/0GRjF.jpg
Plot of the PDF: https://i.stack.imgur.com/pvcQG.jpg
what did I do wrong?
The key is in the support interval. Both $t$ and $u-t$ must lie within the support, or otherwise $p(t)p(u-t)=0$. So the bounds for the integral depends on the value given for $u$.
You have $p(x)=\dfrac 1{2\sqrt{3x}}\mathbf 1_{0\leq x\leq 3}$
The the convolution is $$\begin{split}\int_\Bbb R p(t)p(u-t)~\mathsf d t &= \int_\Bbb R \tfrac{1}{12\sqrt{t(u-t)}}\mathbf 1_{0\leq t\leq 3~,~0\leq u-t\leq 3}~\mathsf d t\\&= \int_\Bbb R \tfrac{1}{12\sqrt{t(u-t)}}\mathbf 1_{0\leq t\leq 3~,~ t\leq u~,~ u-3\leq t}~\mathsf d t\\ &= \int_{\max\{0,u-3\}}^{\min\{3,u\}} \tfrac{1}{12\sqrt{t(u-t)}}~\mathbf 1_{0\leq u\leq 6}~\mathsf d t \\ &= \mathbf 1_{0\leq u\leq 3}\int_{0}^{u} \tfrac{1}{12\sqrt{t(u-t)}}~\mathsf d t+\mathbf 1_{3<u\leq 6}\int_{u-3}^{3} \tfrac{1}{12\sqrt{t(u-t)}}~\mathsf d t\end{split}$$