find laplace transform
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your solution is attached above ........
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Answer:
Use integral definition of laplace transform to get
∫∞0sinh(at)exp(−st)dt
=limb→∞∫b0sinh(at)exp(−st)dt
By the definition sinh(at)=12(exp(at)−exp(−at))
We can write
=limb→∞12∫b0(exp(at)−exp(−at))exp(−st)dt
=limb→∞12∫b0exp((a−s)t)−exp((−a−s)t)dt
limb→∞12[(1a−sexp((a−s)b)−1−a−sexp((−a−s)b))−(1a−s−1−a−s)]
But taking the limit as b→∞ yields infinity terms.
PLEASE REFER TO THE ATTACHMENT ALSO!!
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