• Given,$L[J_0(t)]=\dfrac{1}{\sqrt{s^2+1}}$
• From the given integral, by change of scale property,

$ L[J_0(8t)]=\dfrac{1}{8}.\dfrac{1}{\sqrt{\big[\dfrac{s}{8}\big]^2+1}}$

$\therefore L[J_0(8t)]=\dfrac{1}{\sqrt{s^2+64}}$

• By effect of multiplication by 't',

$ L[tJ_0(8t)]=(-1)^1.\frac{\mathrm{d}\dfrac{1}{\sqrt{s^2+64}}}{\mathrm{d}s}$

$\therefore L[tJ_0(8t)]=(-1).\dfrac{(\sqrt{s^2+64})(0)-(1)\big(\dfrac{2s}{2\sqrt{s^2+64}}\big)}{(\sqrt{s^2+64})^2}$

$\therefore L[tJ_0(8t)]=\dfrac{s}{(s^2+64)(\sqrt{s^2+64})}$

$\therefore \int_{0}^{\infty}e^{-6t}tJ_0(8t)dt=\dfrac{s}{(s^2+64)(\sqrt{s^2+64})}$...(By definition)

• Now, substitute s = 6

$\therefore \int_{0}^{\infty}e^{-6t}tJ_0(8t)dt=\dfrac{3}{500}$