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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?

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  • $\begingroup$ Bessel function $\endgroup$ – Elisabeth Nov 11 '18 at 10:57
  • $\begingroup$ That is Laplace of $x^{2m}$. $\endgroup$ – Nosrati Nov 11 '18 at 15:04
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Hint: $${\cal L}(J_0*J_0)={\cal L}^2(J_0)=\left(\dfrac{1}{\sqrt{s^2+1}}\right)^2=\dfrac{1}{s^2+1}$$

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  • $\begingroup$ The result should be sin (t). This is the Laplace transform of J_0 and I thought is it function f. So I am realy confuse how to find f? $\endgroup$ – Elisabeth Nov 11 '18 at 11:23
  • $\begingroup$ $f$ is $\sin t$, isn't it? $\endgroup$ – Nosrati Nov 11 '18 at 11:27
  • $\begingroup$ Yes, $ f = \sin (t)$. So I should do inverse Laplace transformation afer your hint? $\endgroup$ – Elisabeth Nov 11 '18 at 11:30
  • $\begingroup$ Yes, simply ${\cal L}^{-1}\dfrac{1}{s^2+1}=\sin t$. What is $\cos2x$ in the title? $\endgroup$ – Nosrati Nov 11 '18 at 11:32
  • $\begingroup$ Thank you I add a edit in my question. $\endgroup$ – Elisabeth Nov 11 '18 at 12:47

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