The Bessel's function of the first kind of order zero, $J_0$ is the solution to $$ty''+y'+ty=0$$ which satisfies $J_0(0)=1$
The Laplace transform of this equation gives $$\mathcal{L}[J_0(t)]=\frac{1}{\sqrt{1+s^2}}$$ The problem is to expand $$s\mathcal{L}[J_0(t)]=\frac{s}{\sqrt{1+s^2}}$$ in a binomial series so as to express $J_0$ as a power series.
What I have tried so far: $$s\mathcal{L}[J_0(t)]=\frac{s}{\sqrt{1+s^2}}={\sqrt{\frac{s^2}{1+s^2}}}=\sqrt{1-\frac{1}{1+s^2}}$$ $$\mathcal{L}[J_0(t)]=\frac{1}{s}\sum_{k=0}^\infty \binom{1/2}{k}\left(-\frac{1}{1+s^2}\right)^k$$ But now if I apply $\mathcal{L}^{-1}$ on both sides I get a convolution and a bunch of weird sines inside the series... I'm guessing my problem is expressing $s/\sqrt{1+s^2}$ in terms of $1/s^k$.
Sorry, I don't know how to use LaTex well; there may be a few errors. One can use the following verion of the binomial series $$(1+1/s^2)^{-1/2}=\sum_{k=1}^{\infty}\frac{(-1)^k{(\frac{1}{2})^\left({k}\right)}}{k!s^{2k}}$$ where ${\frac{1}{2}^\left({k}\right)}$ is the Pochhammer symbol (in this case, the rising factorial). Hence taking the Laplace transform of $\frac{1}{s(1+1/s^2)^1/2}$ gives: $$y(t)=\sum_{k=1}^{\infty}\frac{(-1)^k{(\frac{1}{2})^\left({k}\right)}t^{2k}}{k!(2k)!}$$ By using the definition of the rising factorial, and factorial, you should be able to show that $$\frac{{(\frac{1}{2})^\left({k}\right)}}{(2k)!}=\frac{1}{4^kk!} $$ which gives us a series for $J_0(t)$: $$J_0(t)=\sum_{k=1}^{\infty}\frac{(-1)^k}{k!^2}(\frac{t}{2})^{2k}$$