How to find inverse laplace transform of:
$$\frac{1}{\sqrt{s^2+1}}$$
Discussion:
Further:
$$L^{-1}(\frac{1}{\sqrt{s^2+1}})=F(t)$$
Ideas that came to me were:
To use laplace transform of erf($\sqrt{t}$), but that clearly has the limitation of not having $s^2$.
Convolute F(t) so that sin(x) comes on RHS (but that will lead again to an integral equation if we differentiate it twice)
Differentiate $s^{-1}tan^{-1}(s^{-1})=L[Si(t)]$ to introduce $s^{2}+1$, but i see that it would get complicated.
Please help!
Use the binomial series: $$ (1+x)^{-1/2} = \sum_{k=0}^\infty \binom{-1/2}{k}x^k = \sum_{k=0}^\infty = \frac{(-1)^k (2k)!}{2^{2k}(k!)^2}x^k.$$ This gives $$\mathcal{L}\{f(t)\} = \frac{1}{\sqrt{1+s^2}} = \frac{1}{s} \frac{1}{\sqrt{1+\frac{1}{s^2}}} = \frac{1}{s} \sum_{k=0}^\infty \frac{(-1)^k (2k)!}{2^{2k}(k!)^2}\frac{1}{s^{2k}}. $$ Take the inverse Laplace transform for every term: $$ f(t)= \sum_{k=0}^\infty \frac{(-1)^k (2k)!}{2^{2k}(k!)^2}\mathcal{L}^{-1}\left\{\frac{1}{s^{2k+1}}\right\} =\sum_{k=0}^\infty \frac{(-1)^k t^{2k}}{2^{2k}(k!)^2}.$$ This is the power series of the Bessel function $J_0$ of order 0.