 Solution

ID: past-exam-of-the-algebra-analysis-and-differential-equations-1-course-of-the-university-of-oxford/2008/9/iii/solution

Using

sin t = (e^{i t} - e^{- i t}) / (2 i)

and integrating the two exponentials for

Re p > 0

L [sin t] (p) = \frac{1}{2 i} (\frac{1}{p - i} - \frac{1}{p + i}) = \frac{1}{p ^{2} + 1} .

This also follows directly by two integrations by parts in the Laplace transform.

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