 Solution

ID: past-exam-of-the-options-ii-course-of-the-university-of-oxford/2013/d3/a/solution

For

α > 0

[0, R]

gives

\int_{0}^{R} x e^{- αx} d x = \frac{1}{α ^{2}} - e^{- α R} (\frac{R}{α} + \frac{1}{α ^{2}}) .

The exponential remainder tends to zero. Since the integrand is nonnegative, the monotone convergence theorem applied to truncations identifies the improper integral with the Lebesgue integral. Therefore

\int_{0}^{\infty} x e^{- αx} d x = α^{- 2} .

Its finite value proves integrability as well as evaluating it.

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