# Laplace Transform

• Jul 15th 2007, 05:05 PM
fudawala
Laplace Transform
This problem states using integration by parts to find the Laplace transform of the given function; n is a positive integer and a is a real constant:

$
t^2 sin at
$
• Jul 15th 2007, 11:22 PM
red_dog
$\displaystyle\int t^2\sin atdt=\int t^2\left(-\frac{\cos at}{a}\right)'dt=$
$\displaystyle =-\frac{1}{a}t^2\cos at+\frac{2}{a}\int t\cos atdt=$
$\displaystyle =-\frac{1}{a}t^2\cos at+\frac{2}{a}\int t\left(\frac{\sin at}{a}\right)'dt=$
$\displaystyle =-\frac{1}{a}t^2\cos at+\frac{2}{a^2}t\sin at-\frac{2}{a^2}\int\sin atdt=$
$\displaystyle =-\frac{1}{a}t^2\cos at+\frac{2}{a^2}t\sin at+\frac{2}{a^3}\cos at+C$
• Jul 16th 2007, 05:10 AM
CaptainBlack
Quote:

Originally Posted by red_dog
$\displaystyle\int t^2\sin atdt=\int t^2\left(-\frac{\cos at}{a}\right)'dt=$
$\displaystyle =-\frac{1}{a}t^2\cos at+\frac{2}{a}\int t\cos atdt=$
$\displaystyle =-\frac{1}{a}t^2\cos at+\frac{2}{a}\int t\left(\frac{\sin at}{a}\right)'dt=$
$\displaystyle =-\frac{1}{a}t^2\cos at+\frac{2}{a^2}t\sin at-\frac{2}{a^2}\int\sin atdt=$
$\displaystyle =-\frac{1}{a}t^2\cos at+\frac{2}{a^2}t\sin at+\frac{2}{a^3}\cos at+C$

He is asking for the Laplace transform of $t^2 \sin(at)$ so the
integral in question is:

$
\int_0^{\infty} t^2 \sin(at) e^{-st}dt
$

RonL