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

A/s3+B/s7+C/s11+...

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)=

x-x³/3!+x⁵/5!-...(-1)ⁿx²ⁿ⁺¹/(2n+1)!

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:

ℒ{sin(t²)}=∑(-1)ⁿ/(2n+1)!ℒ{t⁴ⁿ⁺²}=∑((-1)ⁿ/(2n+1)!)(4n+2)!/s⁴ⁿ⁺³.

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

2!/s³-(6!/3!)/s⁷+(10!/5!)/s¹¹-(14!/7!)/s¹⁵+...

2/s³-120/s⁷+30240/s¹¹-17297280/s¹⁵+...

You didn’t mess up!

by Top Rated User (839k points)

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