Using the Taylor series,  the Laplace transform of sin(t^2) is 


I'm getting A as 2, B as -120, and C as 30240

I'm not sure where/if I am messing up. 

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1 Answer

Taylor expansion of sin(x)=


Therefore Taylor expansion of sin(t²)=

t²-t⁶/3!+t¹⁰/5!-...(-1)ⁿt⁴ⁿ⁺²/(2n+1)! for integer n≥0.

Now apply Laplace Transform to the series, ℒ{sin(t²)}.

ℒ(tʳ)=r!/sʳ⁺¹; let r=4n+2, then:


So let’s see what the Laplace series looks like:



You didn’t mess up!

by Top Rated User (839k points)

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