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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 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.

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. Last edited by skipjack; January 31st, 2019 at 03:25 PM. 
January 31st, 2019, 11:36 AM  #6  
Newbie Joined: Jan 2019 From: Italy Posts: 2 Thanks: 0  Quote:
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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