 Solution

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

Write

I_{n} (p) = \int_{0}^{\infty} e^{- pt} t^{n} d t

. For

Re p > 0

, exponential decay dominates every fixed power of

t

, so the improper integral converges absolutely and

e^{- pT} T^{n} \to 0

. For

n \geq 1

I_{n} (p) = [- \frac{e ^{- pt} t ^{n}}{p}]_{0}^{\infty} + \frac{n}{p} \int_{0}^{\infty} e^{- pt} t^{n - 1} d t = \frac{n}{p} I_{n - 1} (p) .

The preceding part gives the base case

I_{0} (p) = 1/ p

. Repeated application, or mathematical induction, yields

L {t^{n}} (p) = \frac{n !}{p ^{n + 1}}, Re p > 0.

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