 Solution

ID: past-exam-of-the-options-ii-course-of-the-university-of-oxford/2012/d/d3/c/solution

For every real

x

, the numerator is at most

2 e^{∣ x ∣}

, while the denominator is at least

e^{2∣ x ∣}

. Therefore

0 \leq \frac{e ^{x} + e ^{- x}}{e ^{2 x} + e ^{- 2 x}} \leq 2 e^{- ∣ x ∣} .

The upper bound is integrable by part (a), so the comparison theorem shows that the function is integrable on $R$ .

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