I want to do this integral $H(\rho)=\int_{0}^{\infty} J_1(2 \pi Lr)J_0(2\pi \rho r)dr$, where $J_1$ and $J_0$ are Bessel functions of the first kind and $L\in \mathbb{R}$ is a constant, so I tried to do this in the Mathematica, but he failed. When I tried to put some value to $L$ and $\rho$, the software calculate numerically, so I ploted $H(\rho)$ for a fixed $L$ and the result of the plot is a function like $rect(x/L)$, such that
\begin{equation} {\displaystyle \operatorname {rect} (t)=\left\{{\begin{array}{rl}0,&{\text{if }}|t|>{\frac {1}{2}}\\{\frac {1}{2}},&{\text{if }}|t|={\frac {1}{2}}\\1,&{\text{if }}|t|<{\frac {1}{2}}.\end{array}}\right.} \end{equation}
I'm not sure about this result for $H(\rho)$, so I searched in the internet and didn't find none property for solve this integral, I don't know if, in fact $H(\rho)=rect(x/L)$ or something of this type. Someone knows if this result is correct? This integral have an analytic solution?
The general relationship you need is Eqn. 10.22.63 in the Digital Library of Mathematical Functions: $$ \int_{0}^{\infty}J_{\mu}\left(ax\right)J_{\mu-1}\left(bx\right)\mathrm{d}x=% \begin{cases}b^{\mu-1}a^{-\mu},&0<b<a,\\ (2b)^{-1},&b=a(>0),\\ 0,&0<a<b,\end{cases} $$ assuming $\Re(\mu) > 0$.
In your case, $a = 2 \pi L$, $b = 2 \pi \rho$, and $\mu = 1$. The result you found above is what you'd get if you set $a = 1$ and $b = t$.
While the given equation seems to require positive $a$ and $b$, I believe you can extend the result to all real non-zero $a$ and $b$ by changing the variable of integration $t \to - t$ and/or using the property that $J_n(-x) = (-1)^n J_n(x)$.