= Solution
{c}
Use the causal <convolution> $(f_0*f_1)(x)=\int_0^x f_0(t)f_1(x-t)\,dt$. The <convolution theorem for Laplace transforms> and (i) give
$$
\mathcal L(f_0*f_1)(p)=F_0(p)F_1(p)=\frac1{\sqrt{p^2+1}}-\frac p{p^2+1}.
$$
Since $\mathcal L(\cos x)=p/(p^2+1)$, uniqueness of the <Laplace transform> yields
$$
\boxed{(f_0*f_1)(x)=f_0(x)-\cos x,\qquad x\ge0.}
$$
At $x=0$ the right side is $1-1=0$, agreeing with the empty <convolution> integral. Identification of $f_0$ as a <Bessel function> is unnecessary.
Back to article page
