I have been trying to find out the inverse Laplace transform of 1/(s^2+1)^3 by ways of Convolution Theorem without any success.

I started by identifying F(s) = 1/(s^2+1) => f(t) = Sin(t) and G(s) = 1/(s^2+1)^2 => g(t) = (1/2)[sin(t) - t Cos(t)] and then integrating

"Sin(t-u)[(1/2){Sin(u)-(t-u)Cos(t-u)}] wrt u from u = 0 to u = t. This is where I met my match. I had tried the other way around with no success. Hence, this email. Any suggestion or wisdom? Thank you very much in advance for your guidance. Best regards.

If we will took the integral one at x and the other at u-x, then it is called convolution theorem in mathematics.

Click herehttps://chuksdump.com/appliance-removal-services-in-lynnwood-waAppliance Removal Services in Lynnwood WAA convolution is described mathematically as the integral of two functions, one at x and the other at u-x, across all of space.

Leadership Coaching in DubaiGood morning,

This is an update. On Friday night, 03/24/23, I was able to resole the problem I had posted. By multiplying both the numerator and denominator by "s", I changed F(s) = 1/s(s^2+1) = 1/s - s/s^2+1 => f(t) = 1-cos(t) and G(s) = s/(s^2+1)^2 => g(t) = t sin(t)/2 and successfully integrated (1-cos(t-u))[u sin(u)/2] from u=0 to u=t. The final answer is (1/8){[(3-t^2)sin(t) - 3[tcos(t)]}.

Thank you for your interest in this issue.

Best regards.

Peter Chiang