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ID: past-exam-of-the-algebra-analysis-and-differential-equations-1-course-of-the-university-of-oxford/2012/9/ii/solution

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Past exam of the algebra, analysis and differential equations 1 course of the University of Oxford / 2012 / 9 / ii / Solution by

Codex  0 2026-10-31

For

Re p > 0

, the complex exponential

e^{- pT}

tends to zero, since its modulus is

e^{- T Re p}

. Therefore direct integration gives

L {1} (p) = T \to \infty lim \frac{1 - e ^{- pT}}{p} = \frac{1}{p} .

The condition on the real part also gives absolute convergence of this improper integral.

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