My Math Forum Laplace transform dilemma

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 January 20th, 2019, 11:22 AM #1 Newbie   Joined: Jan 2019 From: Italy Posts: 2 Thanks: 1 Math Focus: Complex Analysis Laplace transform dilemma I was explicitly asked to find the Laplace transform of the following function: $\displaystyle f(t)=\frac{\sin(\omega t)}{\cos^2(\omega t) +1}$ There's no way I can solve it manually with my current abilities. I've tried to run it on Mathematica but it can't find a solution either. By the comparison test for improper integrals, I have concluded that the Laplace transform of $\displaystyle f(t)$ does actually exist. What do you think? Thank you in advance Thanks from topsquark Last edited by skipjack; January 20th, 2019 at 01:41 PM. Reason: symbolism error
 January 20th, 2019, 12:53 PM #2 Global Moderator   Joined: May 2007 Posts: 6,823 Thanks: 723 Since $|f(t)|\le 1$ for all $t$, Laplace transform exists. Thanks from topsquark and Absynthe
 January 20th, 2019, 01:19 PM #3 Senior Member   Joined: Dec 2015 From: somewhere Posts: 645 Thanks: 92 Use integration by parts, $\displaystyle u=e^{-wt}$ and $\displaystyle dv=-\frac{d\cos (wt)}{1+\cos^2 (wt)}$ . Last edited by skipjack; January 20th, 2019 at 01:42 PM.
 January 31st, 2019, 10:22 AM #4 Newbie   Joined: Jan 2019 From: Italy Posts: 2 Thanks: 0 I tried this way but then I do not know how to calculate the laplace transform of L[arctan(cos(ωt))] : do you have any suggestions about it? L[sin(ωt)/(1+cos²(ωt)] = ∫_0,∞ e^(-st)[sin(ωt)/(1+cos²(ωt)] dt ∫e^(-st)[sin(ωt)/(1+cos²(ωt)] dt = (1/ω)∙∫e^(-st)[d(-cos(ωt)) / (1+cos²(ωt)] = - (1/ω)∙∫e^(-st)[d(cos(ωt)) / (1+cos²(ωt)] = - (1/ω)∙∫e^(-st) d [ arctan(cos(ωt)) ] = - (1/ω)∙e^(-st) arctan(cos(ωt)) +(1/ω)∙∫arctan(cos(ωt)) d e^(-st) = - (1/ω)∙e^(-st) arctan(cos(ωt)) -(s/ω)∙∫arctan(cos(ωt)) e^(-st) dt L[sin(ωt)/(1+cos²(ωt)] = (1/ω)∙(π/4) - (s/ω)∙L[arctan(cos(ωt))] Best regards
 January 31st, 2019, 11:17 AM #5 Senior Member   Joined: Dec 2015 From: somewhere Posts: 645 Thanks: 92 Rog are you absynthe ? However . The way I used integration by parts leads to the easiest possible form of integral. Once there is no standard antiderivative, then series and approximations are included. Thanks from Rog Last edited by skipjack; January 31st, 2019 at 03:25 PM.
January 31st, 2019, 11:36 AM   #6
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Quote:
 Originally Posted by idontknow Rog are you absynthe ? However . The way I used integration by parts leads to the easiest possible form of integral. Once there is no standard antiderivative, then series and approximations are included.
I'm not Absynthe, but I'm also interested in solving the problem.

Regarding the resolution with the series, can you give me some further suggestion?

Best regards

Last edited by skipjack; January 31st, 2019 at 03:26 PM.

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