Dear All,
Does anyone have an idea on how to evaluate (or final expression) the Laplace Transform of Arctan(t), i.e. tan^(-1) t? The Laplace Transform of its first derivative 1/(1 + t^2) is also useful. We can deduce the Laplace of the original function accordingly.
Please do not refer me to Laplace transform tables or any other link UNLESS it has the solution to my question. I have searched many references but no luck.
Hope you can provide a help in this issue...
Kind Regards
Dear Hady,
according to Wolfram Mathematica the laplace transform of Arctan(t) reads
(2 CI(s) Sin(s) + Cos(s) (Pi - 2 SI(s) )/(2s)
and the laplace transform of 1/(1 + t^2) evaluates to
CI(s) Sin(s) + 0.5 Cos(s) ( Pi - 2 Si(s))
where CI(s) is the cosine integral function and Si(s) the sine integral function.
You can easiliy reproduce these results by typing
"LaplaceTransform[Arctan[t],t,s]" into the input field at http://www.wolframalpha.com
Hope I could help,
Best Regards
Thank you Peter for your comprehensive answer. In the future I would suggest Hady to consult the tables of Laplace transform. They can be found on the Internet, as Peter did or, old fashioned way, in the University library.
Best regards
Thank you Peter, Arcady and Demetris for your help...
I got the answer I was looking for. Also, I got to learn about something called "auxiliary functions" that can be expressed in terms of trigonometric integrals. The Laplace transform of 1/(1 + t^2) is actually the auxiliary function f(s).
You can read more about this topic in Abramowitz and Stegun (P. 232).
Thanks again for your help...
Good luck
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