# Inverse Laplace transformation - Bessel function

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?

• Bessel function – Elisabeth Nov 11 '18 at 10:57
• That is Laplace of $x^{2m}$. – Nosrati Nov 11 '18 at 15:04

Hint: $${\cal L}(J_0*J_0)={\cal L}^2(J_0)=\left(\dfrac{1}{\sqrt{s^2+1}}\right)^2=\dfrac{1}{s^2+1}$$
• $f$ is $\sin t$, isn't it? – Nosrati Nov 11 '18 at 11:27
• Yes, $f = \sin (t)$. So I should do inverse Laplace transformation afer your hint? – Elisabeth Nov 11 '18 at 11:30
• Yes, simply ${\cal L}^{-1}\dfrac{1}{s^2+1}=\sin t$. What is $\cos2x$ in the title? – Nosrati Nov 11 '18 at 11:32