L { Integral (t- T)^2 Cos(2T) dT }

T = tau

The integral goes from 0 to t

I am unsure how to get to the answer, which is 2/ (s^2)(s^2+4)

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- Apr 4th 2010, 11:59 AMnaiknLaplace Transform : Convolution
L { Integral (t- T)^2 Cos(2T) dT }

T = tau

The integral goes from 0 to t

I am unsure how to get to the answer, which is 2/ (s^2)(s^2+4) - Apr 4th 2010, 01:00 PMchisigma
Given and two functions and and their L-trasforms. The convolution is defined as...

(1)

Is...

(2)

Kind regards