Laplace Transform

**fudawala** · July 15th 2007, 05:05 PM

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:

$<br /> t^2 sin at<br />$

**red_dog** · July 15th 2007, 11:22 PM

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

**CaptainBlack** · July 16th 2007, 05:10 AM

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:

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

RonL

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