\int_{0}^\infty {\cos(xt)\over 1+t^2}dt I'm supposed to solve this using Laplace Transformations.

Jazmine Sweeney 2022-04-30 Answered
0{cos(xt)over1+t2}dt
I'm supposed to solve this using Laplace Transformations.
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Answers (1)

Salvador Ayala
Answered 2022-05-01 Author has 10 answers
Let f(x) denote the integral, and assume temporarily that x>0. This makes no difference since f(x) is even by definition. Then its Laplace transform Lf(s) defines a continuous function on s(0,) (in fact, it defines an analytic function on R(s)>0). Thus we may assume further that s1 and then rely on the continuity argument. Then
Lf(s)=0f(x)esxdx
=00cos(xt)1+t2esxdtdx
=00cos(xt)1+t2esxdxdt
=011+t2ss2+t2dt
=s1s20(1s2+t211+t2)dt
=s1s2π2(1s1)=π211+s
Here, the change of order of integration (∗) is justified by the dominated convergence theorem. Though we proved this for s1, it remains valid by continuity argument as mentioned above. Then by the uniqueness of the Laplace transform, we find that
f(x)=π2ex
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