How to find $f$ using Laplace transformation?
$f = J_0 * J_0$ where * is a convolution. According to the Convolution theorem it is $$(J_0 * J_0)(t):= \int_0^t J_0 (t - \tau) J_0 (\tau)\mathop{\mathrm d \tau}$$
$$J_\nu(z)=\left(\frac{z}{2}\right)^\nu\sum_{k=0}^{\infty}\frac{(-1)^k}{k!\Gamma(\nu+k+1)}\left(\frac{z}{2}\right)^{2k}$$
EDIT Please, can you explain me the equality in the picture? picture And I do not understant why (2m)! is divided by $s^{2m+1}$. There is used Laplace transform of $t^{\alpha}$? Why?
Hint: $${\cal L}(J_0*J_0)={\cal L}^2(J_0)=\left(\dfrac{1}{\sqrt{s^2+1}}\right)^2=\dfrac{1}{s^2+1}$$