144Chapter 8The Laplace and Hankel Transforms with Applications49.We havesint=∞±k=0(-1)kt2k+1(2k+ 1)!for allt.So fort±= 0, we divide bytboth sides and getsintt=∞±k=0(-1)kt2k(2k+ 1)!for allt±=0.Ast→0, the left side approaches 1. The right side is continuous, and so ast→0,it approaches the value at 0, which is 1. Hence both sides of the equality approach1ast→0, and so we may take the expansion to be valid for allt. Apply the resultof the previous exercise, thenL(f(t))(s)=L²∞±k=0(-1)kt2k(2k+ 1)!³(s)=∞±k=0(-1)k(2k)!1s2k+1(2k+ 1)!=∞±k=0(-1)k1s2k+1(2k+1).Recall the expansion of the inverse tangent:tan-1u=∞±k=0(-1)ku2k+12k+1|u|<1.Sotan-11s=∞±k=0(-1)k1s2k+1(2k|s|>1.Comparing series, we Fnd that, fors>1,L(sintt) = tan-1(1s).
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