 Solution

ID: past-exam-of-the-integration-course-of-the-university-of-oxford/2017/2/b/iii/solution

\frac{e ^{- x^{2}}}{x} = \sum_{n = 0}^{\infty} a_{n} (x), a_{n} (x) = \frac{( - 1 ) ^{n} x ^{2 n - 1/2}}{n !}, 0 < x < 1.

\int_{0}^{1} ∣ a_{n} (x) ∣ d x = \frac{1}{n ! ( 2 n + 1/2 )} = \frac{2}{( 4 n + 1 ) n !} .

The sum of these nonnegative integrals is at most

2 \sum_{n \geq 0} 1/ n! = 2 e < \infty

. Applying integration of a series with summable L1 norms, just proved in part (ii), therefore justifies exchanging the series and the integral:

\int_{0}^{1} \frac{e ^{- x^{2}}}{x} d x = 2 n = 0 \sum \infty \frac{( - 1 ) ^{n}}{( 4 n + 1 ) n !} .

The absolute-integral bound is the justification; pointwise convergence of the exponential series alone would not suffice.

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