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    Laplace Transformations

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    a) L [ Sin t / t ]

    L[ Sin t ] = 1/(s^2+1)

    Now, if f(t) <-----> F(s)
    then, f(t)/t <-----> Integral (from s to Infinity) [F(s)ds]

    So, L [ Sin t / t ] = Integral (from s to Infinity) [ds/(s^2+1)]
    L [ Sin t / t ] = tan^-1 (1/s)...( "tan ...

    Solution Summary

    There area variety of Laplace transformations in this solution