Chapter 5.2, Problem 28P

The Laplace transforms of certain functions can be found conveniently from their Taylor series expansions.

(a) Using the Taylor series for $\sin t$,

$\sin t={\displaystyle \underset{n=0}{\overset{\infty }{\sum }}\frac{{\left(-1\right)}^n{t}^{2n+1}}{\left(2n+1\right)!},}$

and assuming that the Laplace transform of this series can be computed term by term, verify that

$L\left\{\sin t\right\}=\frac{1}{s^2+1},s>1$.

(b) Let

$f\left(t\right)=\{\begin{array}{l}\raisebox{1ex}{$\sin t$}\!\left/ \!\raisebox{-1ex}{$t$}\right.,t\ne 0,\\ 1,t=0\end{array}$.

Find the Taylor series for $f$ about $t=0$. Assuming that the Laplace transform of this function can be computed term by term, verify that

$L\left\{f\left(t\right)\right\}=\arctan \left(\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$s$}\right.\right),s>1$.

(c) The Bessel function of the first kind of order zero, $J_0$, has the Taylor series

$J_0\left(t\right)={\displaystyle \underset{n=0}{\overset{\infty }{\sum }}\frac{{\left(-1\right)}^n{t}^{2n}}{2^{2n}{\left(n!\right)}^2}}$.

Assuming that the following Laplace transforms can be computed term by term, verify that

$L\left\{J_0\left(t\right)\right\}={\left(s^2+1\right)}^{-\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.},s>1$,

And

$L\left\{J_0\left(\sqrt{t}\right)\right\}=s^{-1}{e}^{-\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$4s$}\right.},s>0$.