I have the following problem:
I want to compute the value of $<P_T>$ as a function of $P_L$ meaning $$ <P_T> = f(P_L) $$
Where $P_T = Psin(\theta)$ and $P_L = Pcos(\theta)$. P has a uniform distribution in [0, 10] and so do $\theta$ in [0, 2$\pi$].
For simplicity let's call $P\rightarrow x$: $$f_X(x) = \begin{cases}\frac{1}{10} & x \in [0, 10] \\ 0 & otherwise \end{cases} $$
Now:
$$f_\theta(\Theta) = \begin{cases}\frac{1}{2\pi} & \Theta \in [0, 2\pi] \\ 0 & otherwise \end{cases} $$
Let's now call $sin(\theta) \rightarrow y$ then, thanks to Jacobian formula:
$$f_Y(y) = \begin{cases}\frac{1}{\pi\sqrt{(1-y^2)}} & y \in [-1,1] \\ 0 & otherwise \end{cases} $$
I want to compute the pdf of $P_T \rightarrow z = xy$ using the formula: $$f_Z(z) = \int_{-\infty}^{\infty} f_X(x)f_Y(\frac{z}{x})\frac{1}{|x|}dx $$
This means to me: $$f_Z(z) = \int_{0}^{10} \frac{1}{10\pi}\frac{1}{\sqrt{(1-\frac{z^2}{x^2}})}\frac{1}{|x|}dx = \frac{1}{10\pi} \big[log(\sqrt{x^2-z^2} + x)\big]_{x = 0}^{x = 10}$$
Now I have a problem meaning that for $x = 0$ the right hand side of the previous equation evaluates to: $$log(\sqrt{x^2-z^2} + x) |_{x=0} = log(\sqrt{-z^2}) $$
As $z=P_T$ takes real values in the range [0,10] I think that the formula is wrong but I can't find a way to make it right. Can someone spot the error?
BTW to compute $<z>$ I thought of doing $<z> = \int zf_Z(z)dz$ hoping to find some dependence only on $P_L$. That's why, with this reasoning I need the pdf of $z$.
Thank you everyone for your help.