Calculating the Laplace transform of an arbitrary Bessel Function of the first kind

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$t^2 \dfrac {d^2 x} {d t^2} + t \dfrac{d x}{d t}+(t^2-\alpha^2)x=0, \quad \alpha \in \mathbb{C}$

which yields the following result

$\displaystyle J_{\alpha}(t)=\sum_{n=0}^{\infty} \frac{(-1)^n}{n!\Gamma (n+\alpha+1)} \left( \frac{t}{\alpha}\right)^{2n+\alpha} $

Taking the Laplace transform of this function is extremely laborious, so we'll take a different approach, namely, by taking the Laplace transform of the Bessel equation. Therefore we have

$\LL \left[t^2 \dfrac{d^2 x}{d t^2} + t \dfrac{d x}{d t}+(t^2-\alpha^2)x\right](s)=0 \Leftrightarrow \dfrac{d^2}{ds^2}\mathcal{L}[x'']-\dfrac{d}{ds}\mathcal{L}[x']+\dfrac{d^2}{ds^2}\mathcal{L}[x]-\alpha^2\mathcal{L}[x]=0$

We set the initial conditions as $x(0)=1,\ x'(0)=0$, and we get

$\dfrac{d^2}{ds^2}(s^2\mathcal{L}[x]-s)-\dfrac{d}{ds}(s\mathcal{L}[x]-1)+\dfrac{d^2}{ds^2}\mathcal{L}[x]-\alpha^2\mathcal{L}[x] = 0 \Leftrightarrow \\ \Leftrightarrow 2\mathcal{L}[x]+4s\mathcal{L}'[x]+s^2\mathcal{L}''[x] -\mathcal{L}[x]-s\mathcal{L}'[x]+\mathcal{L}''[x]-\alpha^2\mathcal{L}[x] = 0 \Leftrightarrow \\ \Leftrightarrow (s^2+1)\mathcal{L}''[x]+3s\mathcal{L}'[x]+(1-\alpha^2)\mathcal{L}[x]=0$

Making the following change of variable $u=\sqrt{s^2+1}\mathcal{L}[x]$, simplifies the equation to

$\displaystyle u''\sqrt{s^2+1}+\dfrac{s}{\sqrt{s^2+1}}u' =\dfrac{\alpha^2u}{\sqrt{s^2+1}}\Leftrightarrow (u'\sqrt{s^2+1})'=\dfrac{\alpha^2u}{\sqrt{s^2+1}} \Leftrightarrow\\ \Leftrightarrow u'\sqrt{s^2+1}(u'\sqrt{s^2+1})' =\alpha^2u'u \Leftrightarrow \dfrac{1}{2}((u'\sqrt{s^2+1})^2)'=\dfrac{1}{2}\alpha^2(u^2)' \Leftrightarrow \\ \displaystyle \Leftrightarrow \sqrt{s^2+1}u' =-\alpha u \Leftrightarrow \int \dfrac{1}{u}du= \int -\dfrac{\alpha}{\sqrt{s^2+1}}ds\Leftrightarrow \ln(u)=\alpha\ln(\sqrt{s^2+1}-s) \Leftrightarrow \\ \Leftrightarrow u =(\sqrt{s^2+1}-s)^{\alpha}$

Reverting the substitution and remembering that $x(t)$ is the solution to Bessel equation, we conclude that

$\mathcal{L}[J_{\alpha}(t)](s)=\dfrac{(\sqrt{s^2+1}-s)^{\alpha}}{\sqrt{s^2+1}}$