Question

Laplace transform of t²

Answer 1

$F(p) = \int\limits_0^\infty {f(t){e^{ - pt}}} dt = \int\limits_0^\infty {{t^2}{e^{ - pt}}} dt = \left| {\begin{matrix} {u = {t^2}}&{dv = {e^{ - pt}}dt}\\ {du = 2tdt}&{v = - \frac{1}{p}{e^{ - pt}}} \end{matrix}} \right| = \mathop {\lim }\limits_{x \to \infty } \left( { - \left. {\frac{{{t^2}}}{p}{e^{ - pt}}} \right|_0^x + \frac{2}{p}\int\limits_0^x {t{e^{ - pt}}} dt} \right) = \left| {\begin{matrix} {u = t}&{dv = {e^{ - pt}}dt}\\ {du = dt}&{v = - \frac{1}{p}{e^{ - pt}}} \end{matrix}} \right| = \mathop {\lim }\limits_{x \to \infty } \left( { - \left. {\frac{{{t^2}}}{p}{e^{ - pt}}} \right|_0^x + \frac{2}{p}\left( { - \left. {\frac{t}{p}{e^{ - pt}}} \right|_0^x + \frac{1}{p}\int\limits_0^x {{e^{ - pt}}} dt} \right)} \right) = \mathop {\lim }\limits_{x \to \infty } \left( { - \left. {\frac{{{t^2}}}{p}{e^{ - pt}}} \right|_0^x + \frac{2}{p}\left( { - \left. {\frac{t}{p}{e^{ - pt}}} \right|_0^x - \frac{1}{{{p^2}}}\left. {{e^{ - pt}}} \right|_0^x} \right)} \right) = \mathop {\lim }\limits_{x \to \infty } \left( { - \frac{{{x^2}}}{p}{e^{ - px}} + 0 - \frac{{2x}}{{{p^2}}}{e^{ - px}} + 0 - \frac{2}{{{p^3}}}{e^{ - px}} + \frac{2}{{{p^3}}}{e^0}} \right) = 0 + 0 - 0 + 0 - 0 + \frac{2}{{{p^3}}} = \frac{2}{{{p^3}}}$

Answer: $F(p) = \frac{2}{{{p^3}}}$